Topological Spaces

Cesare Auguste Detti (1847-1914) was an Italian painter.

We have sections

General topology
Closed set
Interior and Boundary
Bases and Subbases
Subspaces
Neighborhood System
Continuity
Quotient Space and Quotient Topology
Connected Sets
The Separation Axioms
Convergence Filters
Product Spaces

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

General Topology.

We can recall the concept of open set by referring to the page point set topology in metric space. where we have the following interesting proposition. \begin{equation}{\label{map426}}\tag{1}\mbox{}\end{equation}

Proposition \ref{map426}. Let $(M,d)$ be a metric space.

(i) The intersection of finite collection of open sets in $M$ is an open set in $M$, and the union of any collection of open sets in $M$ is also an open set in $M$.

(ii) The union of finite collection of closed sets in $M$ is a closed set in $M$, and the intersection of any collection of closed sets in $M$ is also a closed set in $M$. $\sharp$

The properties of open sets in metric spaces as shown in Proposition \ref{map426} inspire the following definition.

Definition. A topological space $(X,\tau )$ is a nonempty set $X$ together with a family $\tau$ of subsets of $X$ satisfying the following conditions:

$X\in\tau$ and $\emptyset\in\tau$;
$O_{1}\in\tau$ and $O_{2}\in\tau$ imply $O_{1}\cap O_{2}\in\tau$;
$O_{\gamma}\in\tau$ for $\gamma\in\Gamma$ implies $\bigcup_{\gamma\in\Gamma}O_{\gamma}\in\tau$, where $\Gamma$ is an index set.

The family $\tau$ is called a topology for the set $X$. Each element in $\tau$ is called an open set in $X$. $\sharp$

Example. Let $(X,d)$ be a metric space. By Proposition \ref{map426}, the properties described above are all satisfied by open sets in a metric space $(X,d)$. This says that we can associate a topological space $(X,\tau )$ with the metric space $(X,d)$, where $\tau$ is the family of open sets in $(X,d)$. This topology is called the metric topology induced by $d$. $\sharp$

Example. Given any set $X$ of points, there are always two topologies that can be defined on $X$. One is the trivial topology in which the only open sets are $X$ and $\emptyset$. A second topology is the discrete topology, where every subset of $X$ is an open set in $X$. $\sharp$

Example. The usual topology for $\mathbb{R}$ is the family of all those sets which contain an open interval about each of their points. That is, a subset $A$ of $\mathbb{R}$ is open if and only if, for any $x\in A$, there exists an open interval $(a,b)$ contained in $A$. We can verify that this family of sets is indeed a topology. We also note that an open interval is an open set. $\sharp$

In order to emphasize the topology $\tau$, we can also say that each element in $\tau$ is a $\tau$-open set in $X$. Let $\tau_{1}$ and $\tau_{2}$ be two topologies for $X$. We say that $\tau_{1}$ is smaller than $\tau_{2}$ and $\tau_{2}$ is larger than $\tau_{1}$ when $\tau_{1}\subseteq\tau_{2}$. In other words, $\tau_{1}$ is smaller than $\tau_{2}$ if and only if each $\tau_{1}$-open set is $\tau_{2}$-open. In this case, it is also said that $\tau_{1}$ is coarser than $\tau_{2}$, and that $\tau_{2}$ is finer than $\tau_{1}$.

\begin{equation}{\label{top265}}\tag{2}\mbox{}\end{equation}

Proposition \ref{top265}. Let $\{\tau_{\gamma}\}_{\gamma\in\Gamma}$ be any collection of topologies for $X$. Then $\bigcap_{\gamma\in\Gamma}\tau_{\gamma}$ is also a topology for $X$. $\sharp$

Let ${\cal C}$ be any collection of subsets of $X$. Then, the intersection of all topologies containing ${\cal C}$ is a topology containing ${\cal C}$ by Proposition \ref{top265}. This topology is the coarsest topology such that all of the sets of ${\cal C}$ are open. Then, we obtain the following proposition

Proposition. Let $X$ be a nonempty set, and let ${\cal C}$ be any collection of subsets of $X$. Then, there exists a coarsest topology $\tau$ which contains ${\cal C}$. $\sharp$

Definition. A set $N_{x}$ in a topological space $(X,\tau )$ is a neighborhood of $x\in X$ when there is an open set $O\in\tau$ satisfying $x\in O\subseteq N_{x}$. $\sharp$

We have the following observations.

A neighborhood of a point need not be an open set. However, every open set is a neighborhood of each of its points.
If $\tau$ is the trivial topology, the only neighborhood of a point $x$ is the space $X$.
If $\tau$ is the discrete topology, then every set is a neighborhood of each of its points.
If $\tau$ is the usual topology for $\mathbb{R}$, then a neighborhood of a point is a set containing an open interval to which the point belongs.

\begin{equation}{\label{top7}}\tag{3}\mbox{}\end{equation}

Proposition \ref{top7}. A set is open if and only if it contains a neighborhood of each of its points. Equivalently, a set is open if and only if it is a neighborhood of each of its points

Proof. Suppose that $A$ is open. The definition of neighborhood says that $A$ is a neighborhood of each of its points. Equivalently, the open set $A$ contains a neighborhood of each of its points. For the converse, suppose that $A$ contains a neighborhood of each of its points. Let $U$ be the union all open subsets of $A$. Then $U\subseteq A$ and $U$ is open. For each $x\in A$, the assumption says that there exists a neighborhood $N$ of $x$ satisfying $x\in N\subseteq A$. Since there also exists an open set $O$ containing $x$ satisfying \[x\in O\subseteq N\subseteq A,\] it follows $x\in U$. This shows $U=A$. Therefore $A$ is open, and the proof is complete. $\blacksquare$

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Closed Sets.

Definition. A subset $A$ of a topological space $(X,\tau )$ is said to be closed when its complement set $X\setminus A$ is open. $\sharp$

We have the following observations.

A set is open if and only if its complement is closed.
If $\tau$ is the trivial topology, the complement of $X$ and the complement of the empty set are the only closed sets; that is, only the empty set and $X$ are closed.
It is always true that the space $X$ and the empty set are simultaneously closed and open.
If $\tau$ is the discrete topology, then every subset is simultaneously closed and open.

Example. Let $\tau$ be the usual topology for $\mathbb{R}$. A closed interval is a closed set. An open interval is not a closed set. A half-open interval is neither open nor closed set. The only sets which are both open and closed are the space $\mathbb{R}$ and the empty set. $\sharp$

According to the De Morgan law, the union (resp. intersection) of the complements of the members of a family of sets is the complement of the intersection (resp. union). Therefore, we have the following properties.

The union of a finite number of closed sets is closed.
The intersection of the members of an arbitrary family of closed sets is closed.

Proposition. Let $\tau$ be a family of sets such that the following conditions are satisfied.

The union of a finite subfamily is a member.
The intersection of an arbitrary subfamily is a member.
$X=\bigcup_{F\in\tau}F$ is a member.

Then $\tau$ is precisely the family of closed sets in $X$ relative to the topology consisting of all complements of members of $\tau$. $\sharp$

Definition. The closure of a set $A$ in a topological space $(X,\tau )$ is the intersection of all the members of the family of all closed sets containing $A$.
The closure of $A$ is denoted by $\mbox{cl}(A)$. $\sharp$

We have the following observations.

The set $\mbox{cl}(A)$ is always closed, since it is the intersection of closed sets.
It is obvious that $\mbox{cl}(A)$ is contained in each closed set which contains $A$. This also means that $\mbox{cl}(A)$ is the smallest closed set containing $A$.
We see that $A$ is closed if and only if $A=\mbox{cl}(A)$.
We have the equality $\mbox{cl}(\mbox{cl}(A))=\mbox{cl}(A)$.

Proposition. Let $A$ and $B$ be two sets in a topological space $(X,\tau )$. Then, we have
\[\mbox{cl}(A\cup B)=\mbox{cl}(A)\cup\mbox{cl}(B)\mbox{ and }
\mbox{cl}(A\cap B)\subseteq\mbox{cl}(A)\cap\mbox{cl}(B).\]

Proof. It is clear to see
\[A\subseteq\mbox{cl}(A\cup B)\mbox{ and }B\subseteq\mbox{cl}(A\cup B).\]
Therefore, we have
\[\mbox{cl}(A)\subseteq\mbox{cl}(\mbox{cl}(A\cup B))=\mbox{cl}(A\cup B)\mbox{ and }\mbox{cl}(A)\subseteq\mbox{cl}(A\cup B),\]
which proves the inclusion
\[\mbox{cl}(A)\cup\mbox{cl}(B)\subseteq\mbox{cl}(A\cup B).\]
On the other hand, since $\mbox{cl}(A)\cup\mbox{cl}(B)$ is a closed set containing $A\cup B$, it follows
\[\mbox{cl}(A\cup B)\subseteq\mbox{cl}(A)\cup\mbox{cl}(B).\]

It is clear to see
\[A\cap B\subseteq A\mbox{ and }A\cap B\subseteq B.\]
Therefore, we have
\[\mbox{cl}(A\cap B)\subseteq\mbox{cl}(A)\mbox{ and }\mbox{cl}(A\cap B)\subseteq\mbox{cl}(B).\]
This completes the proof. $\sharp$

Definition. Let $A$ be a subset of a topological space $(X,\tau )$. A point $x\in X$ is called an accumulation point (sometimes called cluster point or limit point) of $A$ when
\[N\cap (A\setminus\{x\})\neq\emptyset\]
for every every neighborhood $N$ of $x$. The set of all accumulation points of $A$ is called the derived set of $A$ and is denoted by $\mbox{der}(A)$ $\sharp$

\begin{equation}{\label{top77}}\tag{4}\mbox{}\end{equation}
Proposition \ref{top77}. Let $A$ be a subset of a topological space $(X,\tau )$. A point $x\in X$ is an accumulation point of $A$ if and only if
\[O\cap (A\setminus\{x\})\neq\emptyset\]
for every open set $O$ containing $x$.

Proof. Let $x$ be an accumulation point of $A$. Then, every neighborhood of $x$ contains points of $A\setminus\{x\}$. Given any open set $O$ containing $x$, Proposition \ref{top7} says that there exists a neighborhood $N$ of $x$ satisfying $x\in N\subseteq O$. Since $N$ contains points of $A\setminus\{x\}$, it follows that the open set $O$ also contains points of $A\setminus\{x\}$.

For the converse, suppose that every open set containing $x$ contains points of $A\setminus\{x\}$. Now, given any neighborhood $N$ of $x$, since there exists an open set $O$ satisfying $x\in O\subseteq N$, it follows that $N$ contains points of $A\setminus\{x\}$. This shows that $x$ is an accumulation point of $A$, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top78}}\tag{5}\mbox{}\end{equation}
Proposition \ref{top78}. Let $A$ be a subset of a topological space. Then, we have
\[\mbox{cl}(A)=A\cup\mbox{der}(A).\]

Proof. We see that, for each neighborhood $N$ of $x$ satisfying $N\cap A\neq\emptyset$ if and only if $x\in A$ or $x\in \mbox{der}(A)$. Suppose that $x\not\in A\cup\mbox{der}(A)$. Then, there exists a neighborhood $N_{x}$ of $x$ satisfying $N_{x}\cap A=\emptyset$. In other words, there exists an open set $O_{x}$ satisfying $x\in O_{x}\subseteq N_{x}$ and $A\cap O_{x}=\emptyset$. Since $O_{x}$ is a neighborhood of each of its points, it means that each point of $O_{x}$ is not an accumulation point of $A$, which says
\[O_{x}\cap (A\cup\mbox{der}(A))=\emptyset.\]
It follows
\[A\cup\mbox{der}(A)=\left (\bigcup_{\scriptsize x\not\in A\cup\mbox{der}(A)}O_{x}\right )^{c}.\]
This shows that $A\cup\mbox{der}(A)$ is closed, which also says that $\mbox{cl}(A)\subseteq A\cup\mbox{der}(A)$, since $\mbox{cl}(A)$ is the smallest closed set containing $A$.

On the other hand, let $B$ be any closed set containing $A$. If $x\in\mbox{der}(A)$, then $x\in\mbox{der}(B)$; that is, every neighborhood $N$ of $x$
contains points of set $B\setminus\{x\}$. Assume that $x\in X\setminus B$. Since $X\setminus B$ is open, there exists a neighborhood $N$ of $x$ satisfying $x\in N\subseteq X\setminus B$, which contradicts $N_{x}$ contains points of set $B\setminus\{x\}$. Therefore, we must have $x\in B$, i.e., $\mbox{der}(A)\subseteq B$. In other words, the derived set $\mbox{der}(A)$ is contained in each closed set containing $A$. This shows $\mbox{der}(A)\subseteq\mbox{cl}(A)$. Therefore, we conclude that $\mbox{cl}(A)$ contains $A\cup\mbox{der}(A)$. This completes the proof. $\blacksquare$

Example. Let $\tau$ be the usual topology for $\mathbb{R}$. Then, we have the following properties.

If $A$ is an open interval $(0,1)$, then every point of the closed interval $[0,1]$ is an accumulation point of $A$.
If $A$ is the set of all nonnegative rational numbers with squares less than $2$, then the closed interval $[0,\sqrt{2}]$ is the set of accumulation points.
If $A$ is the set of all reciprocals of integers, then $0$ is the only accumulation point of $A$.
The set of integers has no accumulation point. $\sharp$

The function which assigns to each subset $A$ of a topological space the value $\mbox{cl}(A)$ might be called the closure operator relative to the topology. This operator determines the topology completely in the sense that a set $A$ is closed if and only if $A=\mbox{cl}(A)$. In other words, the closed sets are simply the sets which are fixed under the closure operator. Four very simple properties serve to describe closure.

The empty set is closed, the closure of the empty set is empty.
Each set is contained in its closure.
Since the closure of each set is closed, the closure of the closure of a set is identical with the closure of the set.
The closure of the union of two sets is the union of the closures, since $\mbox{cl}(A\cup B)$ is always a closed set containing $A$ and $B$, and therefore contains $\mbox{cl}(A)$ and $\mbox{cl}(B)$ and hence $\mbox{cl}(A)\cup \mbox{cl}(B)$. On the other hand, $\mbox{cl}(A)\cup \mbox{cl}(B)$ is a closed set containing $A\cup B$ and hence also $\mbox{cl}(A\cup B)$.

Definition. A closure operator on $X$ is an operator which assigns to each subset $A$ of $X$ a subset $c(A)$ of $X$ such that the following four statements hold true:

if $\emptyset$ is the empty set, then $c(\emptyset )=\emptyset$;
$A\subseteq c(A)$ for each subset $A$;
$c(c(A))=c(A)$ for each subset $A$;
for any subsets $A$ and $B$, $c(A\cup B)=c(A)\cup c(B)$.

The following theorem shows that these four statements are actually characteristic of closure. The topology defined below is the topology associated with a closure operator.

Proposition. Let $c$ be a closure operator on $X$, let $\eta$ be the family of all subsets $A$ of $X$ for which $c(A)=A$, and let $\tau$ be the family of complements of members of $\eta$. Then $\tau$ is a topology for $X$, and $c(A)$ is the $\tau$-closure of $A$ for each subset $A$ of $X$. $\sharp$

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Interior and Boundary.

Definition. Let $A$ be a subset of a topological space $(X,\tau )$. A point $x\in A$ is called an interior point of $A$ when $A$ is a neighborhood of $x$. The set of all interior points of $A$ is the interior of $A$, and is denoted by $\mbox{int}(A)$. $\sharp$

Proposition. Let $A$ be a subset of a topological space $(X,\tau )$. Then, we have the following properties.

(i) The interior $\mbox{int}(A)$ of $A$ is open and is the largest open subset of $A$.

(ii) A set $A$ is open if and only if $A=\mbox{int}(A)$.

(iii) The set of all points of $A$ which are not points of accumulation of $X\setminus A$ is precisely $\mbox{int}(A)$.

(iv) The closure of $X\setminus A$ is $X\setminus \mbox{int}(A)$.

Proof. To prove part (i), if $x\in\mbox{int}(A)$, then there exists an open set $O_{x}$ such that $x\in O_{x}\subseteq A$. We also see that if $y\in O_{x}$, then $y\in\mbox{int}(A)$ by definition, i.e., $O_{x}\subset\mbox{int}(A)$. Therefore, it follows that $\mbox{int}(A)$ contains a neighborhood of each of its
points, which says that $\mbox{int}(A)$ is open by Proposition \ref{top7}. If $O$ is an open subset of $A$ and $x\in O$, then $A$ is a neighborhood of $x$, i.e., $x\in\mbox{int}(A)$, which says that $O\subseteq\mbox{int}(A)$.

To prove part (ii), if $A$ is open, then $A$ is surely identical with the largest open subset of $A$. This says that $A$ is open if and only if $A=\mbox{int}(A)$.

To prove part (iii), if $x$ is a point of $A$ which is not an accumulation point of $X\setminus A$, then there is a neighborhood $N$ of $x$ which does not intersect $X\setminus A$ and is therefore a subset of $A$. Then $A$ is a neighborhood of $x$ and $x\in \mbox{int}(A)$. On the other hand, $\mbox{int}(A)$ is a neighborhood of each of its points and $\mbox{int}(A)$ does not intersect $X\setminus A$, so that no point of $\mbox{int}(A)$ is an accumulation point of $X\setminus A$.

To prove part (iv), since $\mbox{int}(A)$ consists of the points of $A$ which are not accumulation points of $X\setminus A$, the complement, $X\setminus \mbox{int}(A)$, is precisely the set of all points which are either points of $X\setminus A$ or accumulation points of $X\setminus A$; that is, the complement is the closure $\mbox{cl}(X\setminus A)$. This completes the proof. $\blacksquare$

Example. Let $(\mathbb{R},\tau )$ be the set of real numbers with the usual topology.

The interior of the set of all integers is empty.
The interior of a closed interval is the open interval with the same endpoints.
The interior of the set of rational numbers is empty.
The closure of the set of rational numbers is the entire set $\mathbb{R}$ . $\sharp$

Proposition. Let $A$ and $B$ be two subsets of a topological space $(X,\tau )$. Then, we have the following properties.

(i) We have $\mbox{int}(\mbox{int}(A))=\mbox{int}(A)$.

(ii) We have $\mbox{int}(A\cap B)=\mbox{int}(A)\cap\mbox{int}(A)$. $\sharp$

Definition. The boundary of a subset $A$ of a topological space $(X,\tau )$ is the set of all points $x$ which are not the interior points of $A$ and $X\setminus A$. $\sharp$

We have the following observations.

$x$ is a point of the boundary if and only if each neighborhood of $x$ intersects both $A$ and $X\setminus A$.
Since an open set is a neighborhood of each of its points, we see that $x$ is a point of the boundary if and only if each open set containing $x$ intersects both $A$ and $X\setminus A$.
The boundary of $A$ is identical with the boundary of $X\setminus A$.

Example. Let $(\mathbb{R},\tau )$ be the set of real numbers with the usual topology.

The boundary of an interval in $\mathbb{R}$ is the set whose only members are the endpoints of the interval, regardless of whether the interval is open, closed, or half-open.
The boundary of the set of rationals or the set of irrationals is the set of all real numbers. $\sharp$

Proposition. Let $A$ be a subset of a topological space $(X,\tau )$, and let $\mbox{bd}(A)$ be the boundary of $A$.

(i) We have the following equalities.
\begin{align*}
& \mbox{bd}(A)=\mbox{cl}(A)\cap\mbox{cl}(X\setminus A)=\mbox{cl}(A)\setminus\mbox{int}(A)=\mbox{bd}(X\setminus A)\\
& X\setminus\mbox{bd}(A)=\mbox{int}(A)\cup\mbox{int}(X\setminus A)\\
& \mbox{cl}(A)=A\cup\mbox{bd}(A)=\mbox{int}(A)\cup\mbox{bd}(A)\\
& \mbox{int}(A)\cap\mbox{bd}(A)=\emptyset\\
& \mbox{int}(A)=A\setminus\mbox{bd}(A)=\mbox{cl}(A)\setminus\mbox{bd}(A).
\end{align*}

(ii) A set is closed if and only if it contains its boundary.

(iii) A set is open if and only if it is disjoint from its boundary.

Proof. We want to prove $\mbox{bd}(A)=\mbox{bd}(X\setminus A)$. Suppose that $x\in\mbox{bd}(A)$. Then, any open set containing $x$ meets both $A$ and $X\setminus A$. Therefore, $O$ meets $X\setminus A$ and $X\setminus (X\setminus A)=A$, which says $x\in\mbox{bd}(X\setminus A)$. We can similarly show that $x\in\mbox{bd}(X\setminus A)$ implies $x\in\mbox{bd}(A)$. Therefore, we obtain the desired equality.

We want to prove $\mbox{cl}(A)=\mbox{int}(A)\cup\mbox{bd}(A)$. Suppose that $x\in\mbox{cl}(A)$ and $x\not\in\mbox{bd}(A)$. There exists an open set $O$ contains $x$ satisfying $O\subseteq A$ or $O\subseteq X\setminus A$. If $O\subseteq X\setminus A$, then $X\setminus O$ is a closed set containing $A$. This says $\mbox{cl}(A)\subseteq X\setminus O$, which contradicts $x\in\mbox{cl}(A)$. Therefore, we must have $O\subseteq A$, which also says $x\in\mbox{int}(A)$. This shows the inclusion $\mbox{cl}(A)\subseteq\mbox{int}(A)\cup\mbox{bd}(A)$. Now we assume $y\in\mbox{int}(A)\cup\mbox{bd}(A)$ and $y\not\in\mbox{cl}(A)$. Since $\mbox{int}(A)\subset\mbox{cl}(A)$, we must have $y\in\mbox{bd}(A)$. Since $y\not\in\mbox{cl}(A)$, the open set $X\setminus\mbox{cl}(A)$ contains $y$ and does not intersect $A$, which says $y\not\in\mbox{bd}(A)$. This contradiction says $y\in\mbox{cl}(A)$. Therefore, we obtain the inclusion $\mbox{int}(A)\cup\mbox{bd}(A)\subseteq\mbox{cl}(A)$.

We wan to prove $\mbox{int}(A)\cap\mbox{bd}(A)=\emptyset$. Suppose that it is not. For $x\in\mbox{int}(A)\cap\mbox{bd}(A)$, since $x\in\mbox{bd}(A)$, every open set containing $x$ meets $X\setminus A$. However, since $x\in\mbox{int}(A)$, there exists an open set $O$ containing $x$ satisfying $O\subseteq A$, which contradicts that $O$ meets $X\setminus A$. Therefore, we must have $\mbox{int}(A)\cap\mbox{bd}(A)=\emptyset$.

From the facts of $\mbox{int}(A)\cap\mbox{bd}(A)=\emptyset$ and $\mbox{cl}(A)=\mbox{int}(A)\cup\mbox{bd}(A)$, we can also obtain $\mbox{int}(A)=\mbox{cl}(A)\setminus\mbox{bd}(A)$. $\blacksquare$

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Bases and Subbases.

Definition. Let $(X,\tau )$ be a topological space.

A subfamily ${\cal B}$ of $\tau$ is said to be a base for $\tau$ when, for each $x\in X$ and each neighborhood $N$ of $x$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq N$.
A collection ${\cal B}_{x}$ of $\tau$ containing $x$ is called a local base at $x$ when, for each neighborhood $N$ of $x$, there exists $B\in {\cal B}_{x}$ satisfying $x\in B\subseteq N$. $\sharp$

\begin{equation}{\label{top266}}\tag{7}\mbox{}\end{equation}
Proposition \ref{top266}. ${\cal B}$ is a base for $\tau$ if and only if, for each $x\in X$ and each open set $O$ containing $x$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$.

Proof. Since each open set containing $x$ is also a neighborhood of $x$, it follows that if ${\cal B}$ is a base for $\tau$ then, for each $x\in X$ and each open set $O$ containing $x$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$. For the converse, for each $x\in X$ and each neighborhood $N$ of $x$, the exists an open set $O\in\tau$ satisfying $x\in O\subseteq N$. The assumption says that there exists $B\in {\cal B}$ satisfying
\[x\in B\subseteq O\subseteq N.\]
This completes the proof. $\blacksquare$

Example. The collection of all open intervals is a base for the usual topology of $\mathbb{R}$. Let $X$ be a metric space. Then, the set of all open balls forms a base. Also, the set of all open balls centered at $x$ forms a local base at $x$. $\sharp$

\begin{equation}{\label{top270}}\tag{8}\mbox{}\end{equation}
Proposition \ref{top270}. Let ${\cal B}$ be a base for the topology $\tau$. Then $O\in\tau$ if and only if, for each $x\in O$ there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$.

Proof. Suppose that $O\in\tau$ and $x\in O$. Proposition~\ref{top7} says that there exists a neighborhood $N$ of $x$ satisfying $x\in N\subseteq O$. The definition of base says that there exists $B\in {\cal B}$ satisfying $x\in B\subseteq N$. Therefore, we obtain $x\in B\subseteq O$. For the converse, suppose that, for each $x\in O$, we have $x\in B\subseteq O$. Then $O$ must be the union of those $B\in {\cal B}$ satisfying $B\subseteq O$. This says that $O$ is open, since it is a union of open sets. This completes the proof. $\blacksquare$

\begin{equation}{\label{top10}}\tag{9}\mbox{}\end{equation}
Proposition \ref{top10}. A subfamily ${\cal B}$ of $\tau$ is a base for $\tau$ if and only if each open set is the union of members of ${\cal B}$.

Proof. Suppose that ${\cal B}$ is a base for the topology $\tau$, and that $O\in\tau$. Let $U$ be the union of all members of ${\cal B}$ which are subsets of $O$. Then $U\subseteq O$. Given any $x\in O$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$ by Proposition \ref{top270}. Therefore, we have $x\in U$, i.e, $O\subseteq U$. This shows $U=O$. For the converse, suppose that, for $O\in\tau$, $O$ is the union of the members of ${\cal B}$. For each $x\in X$ and each open set $O$ containing $x$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$, which says that ${\cal B}$ is a base for $\tau$ by Proposition \ref{top266}. This completes the proof. $\blacksquare$

\begin{equation}{\label{top267}}\tag{10}\mbox{}\end{equation}
Proposition \ref{top267}. Let ${\cal B}$ be a collection of subsets of a set $X$ such that the following conditions are satisfied.

Each $x\in X$ is contained in some $B\in {\cal B}$. In other words, $X$ is the union of members of ${\cal B}$.
Given any $B_{1},B_{2}\in {\cal B}$ and any $x\in B_{1}\cap B_{2}$, there exists $B_{3}\in {\cal B}$ satisfying $x\in B_{3}\subseteq B_{1}\cap B_{2}$.

Then, the family $\tau$ of subsets of $X$ defined by
\begin{align*}
\tau & =\left\{O:\mbox{given any $x\in O$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$}\right\}\\
& =\left\{O:\mbox{$O$ is the union of members of ${\cal B}$}\right\}.
\end{align*}
is a topology for $X$ such that ${\cal B}$ is a base for $\tau$.

Proof. We consider the axioms of topology.

It is clear to see $\emptyset\in\tau$. The first condition for ${\cal B}$ also says $X\in\tau$.
A union of members of $\tau$ is itself a union of members of ${\cal B}$, which is also a member of $\tau$.
Given any two sets $O_{1}$ and $O_{2}$ in $\tau$, let $x\in O_{1}\cap O_{2}$, i.e., $x\in O_{1}$ and $x\in O_{2}$. The definition of $\tau$ says that there exist $B_{1}$ and $B_{2}$ in ${\cal B}$ satisfying
\[x\in B_{1}\subseteq O_{1}\mbox{ and }x\in B_{2}\subseteq O_{2},
\mbox{ i.e., }x\in B_{1}\cap B_{2}\subseteq O_{1}\cap O_{2}.\]
The second condition says that there exists $B_{3}\in {\cal B}$ satisfying $x\in B_{3}\subseteq B_{1}\cap B_{2}$. It follows $x\in B_{3}\subseteq O_{1}\cap O_{2}$, which shows $O_{1}\cap O_{2}$ is in $\tau$.

The above three cases conclude that $\tau$ is a topology for $X$. Finally, Proposition \ref{top10} says that ${\cal B}$ is also a base for $\tau$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top11}}\tag{11}\mbox{}\end{equation}
Corollary \ref{top11}. Let ${\cal B}$ be a family of sets, and let $X$ be the union of all members of ${\cal B}$. Then ${\cal B}$ is a base for some topological space $(X,\tau )$ if and only if, given any $B_{1},B_{2}\in {\cal B}$ and any $x\in B_{1}\cap B_{2}$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq B_{1}\cap B_{2}$.

Proof. Let ${\cal B}$ be a base for some topology $\tau$ for $X$. Given any $B_{1},B_{2}\in {\cal B}$ and any $x\in B_{1}\cap B_{2}$, since $B_{1}\cap B_{2}\in\tau$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq B_{1}\cap B_{2}$. For the converse, let ${\cal B}$ be a family with the specified property. We define a family $\tau$ by
\[\tau =\left\{O:\mbox{$O$ is the union of members of ${\cal B}$}\right\}.\]
Using Proposition \ref{top267}, we complete the proof. $\blacksquare$

\begin{equation}{\label{top12}}\tag{12}\mbox{}\end{equation}
Corollary \ref{top12}. Let ${\cal S}$ be a family of sets, and let $X$ be the union of all members of ${\cal S}$. The family ${\cal B}$ of all finite intersections of members of ${\cal S}$ is a base for some topological space $(X,\tau )$.

Proof. It is obvious that the intersection of any two members of ${\cal B}$ is also a member of ${\cal B}$. Applying Corollary \ref{top11}, it follows that ${\cal B}$ is a base for some topological space $(X,\tau )$. This completes the proof. $\sharp$

Inspired by Corollary \ref{top12}, we propose the following definition.

Definition. Let $(X,\tau )$ be a topological space, and let ${\cal S}$ be a family of subsets of $X$. We say that ${\cal S}$ is a subbase for $\tau$ when the family of finite intersections of members of ${\cal S}$ is a base for $\tau$. $\sharp$

We see that if ${\cal S}$ is a subbase for $(X,\tau )$ then each member of $\tau$ is the union of finite intersections of members of ${\cal S}$. According to Corollary \ref{top12}, every nonempty family ${\cal S}$ is the subbase for some topology, and this topology is uniquely determined by ${\cal S}$. It is the smallest topology containing ${\cal S}$. In other words, it is a topology containing ${\cal S}$ and is a subfamily of every topology containing ${\cal S}$. In general, there will be many different bases and subbases for a topology and the most appropriate choice may depend on the problem under consideration.

Example. One natural subbase for the usual topology for $\mathbb{R}$ is the family of half-infinite open intervals. More precisely, the family of the form

\[\{x:x>a\}\mbox{ or }\{x:x<a\}.\]

Each open interval is the intersection of two such sets. $\sharp$

We conclude that there are four means of specifying topology on a set $X$:

explicitly giving the open sets; that is, the members of the topology;
explicitly giving the closed sets;
giving a base for the topology;
giving a subbase for the topology.

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Subspaces.

Let $(X,\tau_{X})$ be a topological space, and let $Y$ be a subset of $X$. We may construct a topology $\tau_{Y}$ for $Y$ which is called the relative topology, or the relativization of $\tau_{X}$ to $Y$. The relative topology $\tau_{Y}$ is defined by
\begin{equation}{\label{top14}}\tag{13}
\tau_{Y}=\left\{O\cap Y:O\in\tau_{X}\right\}.
\end{equation}
The following proposition shows that $\tau_{Y}$ is actually a topology for $Y$.

Proposition. The family $\tau_{Y}$ defined in $(\ref{top14})$ forms a topology for $Y$.

Proof. Since $X$ and $\emptyset$ are in $\tau_{X}$, it follows that
\[Y\cap X=Y\mbox{ and }Y\cap\emptyset =\emptyset\]
are in $\tau_{Y}$. Given any $O_{Y}^{(1)},O_{Y}^{(2)}$ in $\tau_{Y}$, there exists $O_{X}^{(1)}$ and $O_{X}^{(2)}$ in $\tau_{X}$ satisfying
\[O_{Y}^{(1)}=Y\cap O_{X}^{(1)}\mbox{ and }O_{Y}^{(2)}=Y\cap O_{X}^{(2)}.\]
Therefore, we obtain
\[O_{Y}^{(1)}\cap O_{Y}^{(2)}=Y\cap\left (O_{X}^{(1)}\cap O_{X}^{(2)}\right ).\]
Since $O_{X}^{(1)}\cap O_{X}^{(2)}\in\tau_{X}$, it follows $O_{Y}^{(1)}\cap O_{Y}^{(2)}\in\tau_{Y}$. Given a family $\{O_{Y}^{(\alpha )}\}_{\alpha\in\Lambda}$ in $\tau_{Y}$, we have $O_{Y}^{(\alpha )}=Y\cap O_{X}^{(\alpha )}$ for some $O_{X}^{(\alpha )}\in\tau_{X}$. Therefore, we obtain
\[\bigcup_{\alpha\in\Lambda}O_{Y}^{(\alpha )}=\bigcup_{\alpha\in\Lambda} \left (Y\cap O_{X}^{(\alpha )}\right )=Y\cap
\left (\bigcup_{\alpha\in\Lambda}O_{X}^{(\alpha )}\right ).\]
Since $\bigcup_{\alpha\in\Lambda}O_{X}^{(\alpha )}\in\tau_{X}$, it follows $\bigcup_{\alpha\in\Lambda}O_{Y}^{(\alpha )}\in\tau_{Y}$. This completes the proof. $\blacksquare$

Example. Let $(X,\tau_{X})$ be a discrete topological space. Then, every subspace of $(X,\tau_{X})$ is also a discrete topological space. Let $(X,\tau_{X})$ be a trivial topological space. Then, every subspace of $(X,\tau_{X})$ is also a trivial topological space. $\sharp$

Each member $O_{Y}$ of the relative topology $\tau_{Y}$ is said to be open in $Y$ or $\tau_{Y}$-open, and its relative complement $Y\setminus O_{Y}$ is closed in $Y$; that is, the relative complement $Y\setminus O_{Y}$ is $\tau_{Y}$-closed. The $\tau_{Y}$-closure of a subset of $Y$ is its closure in} $Y$. Each subset $Y$ of $X$ is both open and closed in itself, although $Y$ may be neither open nor closed in $X$. More formally, we have the following definition.

Definition. Let $Y$ be a subset of a topological space $(X,\tau_{X})$. We define a topology $\tau_{Y}$ by

\[\tau_{Y}=\left\{O\cap Y:O\in\tau_{X}\right\}.\]
We call $\tau_{Y}$ the topology inherited from $\tau_{X}$ and call the topological space $(Y,\tau_{Y})$ a subspace of $(X,\tau_{X})$. $\sharp$

Proposition. Let $(Y,\tau_{Y})$ be a subspace of $(X,\tau_{X})$, and let $(Z,\tau_{Z})$ be a subspace of $(Y,\tau_{Y})$. Suppose that $(Z,\tau_{Z}^{*})$ is also a subspace of $(X,\tau_{X})$. Then, we have $\tau_{Z}=\tau_{Z}^{*}$.

Proof. Given any $O_{Z}\in\tau_{Z}$, there exists $O_{Y}\in\tau_{Y}$ satisfying $O_{Z}=Z\cap O_{Y}$. Since $(Y,\tau_{Y})$ is a subspace of $(X,\tau_{X})$, there exists $O_{X}\in\tau_{X}$ satisfying $O_{Y}=X\cap O_{X}$. Then, we have
\[O_{Z}=Z\cap \left (X\cap O_{X}\right )=(Z\cap X)\cap O_{X}=Z\cap O_{X},\]
which says $O_{Z}\in\tau_{Z}^{*}$. Therefore, we obtain the inclusion $\tau_{Z}\subseteq\tau_{Z}^{*}$. On the other hand, given any $O_{Z}\in\tau_{Z}^{*}$, there exists $O_{X}\in\tau_{X}$ satisfying $O_{Z}=Z\cap O_{X}$. Since $Z\subseteq Y$, we obtain
\[O_{Z}=Z\cap O_{X}=(Y\cap Z)\cap O_{X}=Z\cap (Y\cap O_{X}),\]
which says $O_{Z}\in\tau_{Y}$, since $Y\cap O_{X}\in\tau_{Y}$. Therefore, we obtain the inclusion $\tau_{Z}^{*}\subseteq\tau_{Z}$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top74}}\tag{14}\mbox{}\end{equation}
Proposition \ref{top74}. Let $(Y,\tau_{Y})$ be a subspace of the topological space $(X,\tau_{X})$, and let $O$ be a subset of $Y$. Then, we have the following properties.

(i) Suppose that $O$ is $\tau_{X}$-open. Then $O$ is also $\tau_{Y}$-open. Moreover, we have the inclusions
\[\tau_{X}\mbox{-int}(O)\subseteq\tau_{Y}\mbox{-int}(O)\mbox{ and }\tau_{X}\mbox{-cl}(O)\subseteq\tau_{Y}\mbox{-cl}(O).\]

(ii) Suppose that there exists another subset $K$ of $Y$ satisfying $O\subseteq\tau_{X}\mbox{-int}(K)$. Then $O$ is $\tau_{X}$-open if and only if $O$ is $\tau_{Y}$-open. Moreover, we have the following equalities
\[\tau_{X}\mbox{-int}(O)=\tau_{Y}\mbox{-int}(O)\mbox{ and }\tau_{X}\mbox{-cl}(O)=\tau_{Y}\mbox{-cl}(O).\]
In particular, if $Y$ is $\tau_{X}$-open, then the above properties are satisfied.

Proof. To prove part (i), since $O\subseteq Y$, we have $O=O\cap Y$. Since $O$ is $\tau_{X}$-open, it follows that $O$ is $\tau_{Y}$-open. Given any $x\in\tau_{X}\mbox{-int}(O)$, there exists a $\tau_{X}$-open set $O^{*}$ satisfying
\[x\in O^{*}\subseteq\tau_{X}\mbox{-int}(O)\subseteq O\subseteq Y.\]
Since $O^{*}\subseteq Y$, it says that $O^{*}$ is also $\tau_{Y}$-open as shown above. In other words, there exists a $\tau_{Y}$-open set $O^{*}$ satisfying $x\in O^{*}\subseteq O$, which says $x\in\tau_{Y}\mbox{-int}(O)$. Therefore, we obtain the inclusion
\[\tau_{X}\mbox{-int}(O)\subseteq\tau_{Y}\mbox{-int}(O).\]
Now, given any $x\in\tau_{X}\mbox{-cl}(O)$, Propositions \ref{top77} and \ref{top78} say that either $x\in O$ or $x\not\in O$ and every $\tau_{X}$-open set $O^{*}$ satisfies
\[O^{*}\cap (O\setminus\{x\})\neq\emptyset.\]
Since each $O^{*}$ is also $\tau_{Y}$-open set, it follows $x\in\tau_{Y}\mbox{-cl}(O)$. Therefore, we obtain the inclusion
\[\tau_{X}\mbox{-cl}(O)\subseteq\tau_{Y}\mbox{-cl}(O).\]

To prove part (ii), suppose that $O$ is $\tau_{Y}$-open. Then $O=Y\cap O^{*}$ for some $\tau_{X}$-open set $O^{*}$. Since $O\subseteq\tau_{X}\mbox{-int}(K)$, we have
\[O=\tau_{X}\mbox{-int}(K)\cap O=\tau_{X}\mbox{-int}(K)\cap Y\cap O^{*}=\tau_{X}\mbox{-int}(K)\cap O^{*},\]
which says that $O$ is $\tau_{X}$-open. Moreover the desired equalities can be similarly obtained. This completes the proof. $\blacksquare$

\begin{equation}{\label{top13}}\tag{15}\mbox{}\end{equation}
Proposition \ref{top13}. Let $(X,\tau_{X})$ be a topological space, and let $(Y,\tau_{Y})$ be a subspace of $(X,\tau_{X})$. Then, we have the following properties.

(i) $F_{Y}$ is $\tau_{Y}$-closed in $Y$ if and only if $F_{Y}=Y\cap F_{X}$ for some $\tau_{X}$-closed set $F_{X}$ in $X$.

(ii) Let $F$ be a subset of $Y$. If $F$ is $\tau_{X}$-closed, then $F$ is also $\tau_{Y}$-closed.

(iii) Let $A$ be a subset of $Y$. Then, we have
\[\tau_{Y}\mbox{-cl}(A)=Y\cap\tau_{X}\mbox{-cl}(A).\]
Moreover, if $Y$ is $\tau_{X}$-closed, then we have
\[\tau_{Y}\mbox{-cl}(A)=\tau_{X}\mbox{-cl}(A).\]

Proof. To prove part (i), suppose that $F_{Y}$ is $\tau_{Y}$-closed in $Y$. Since $Y\setminus F_{Y}$ is $\tau_{Y}$-open, there exists a $\tau_{X}$-open set $O_{X}$ in $X$ satisfying $Y\setminus F_{Y}=Y\cap O_{X}$. We want to claim
\[Y\cap (X\setminus O_{X})=Y\setminus (Y\cap O_{X}).\]
Given any $y\in Y\cap (X\setminus O_{X})$, it means $y\in Y$ and $y\not\in O_{X}$, which implies
\[y\in Y\setminus O_{X}\subseteq Y\setminus (Y\cap O_{X}).\]
Therefore, we obtain the inclusion
\[Y\cap (X\setminus O_{X})\subseteq Y\setminus (Y\cap O_{X}).\]
On the other hand, given any $y\in Y\setminus (Y\cap O_{X})$, it means $y\in Y$ and $y\not\in Y\cap O_{X}$, which implies $y\not\in O_{X}$, i.e., $y\in X\setminus O_{X}$. Therefore, we obtain the inclusion
\[Y\setminus (Y\cap O_{X})\subseteq Y\cap (X\setminus O_{X}).\]
Since
\[Y\cap (X\setminus O_{X})=Y\setminus (Y\cap O_{X})=Y\setminus (Y\setminus F_{Y})=F_{Y},\]
we see that $F_{X}\equiv X\setminus O_{X}$ is a $\tau_{X}$-closed set in $X$. For the converse, since
\[Y\setminus F_{Y}=Y\setminus (Y\cap F_{X})=Y\cap (X\setminus F_{X})\]
and $X\setminus F_{X}$ is $\tau_{X}$-open, it follows that $Y\setminus F_{Y}$ is $\tau_{Y}$-open.

To prove part (ii), since $F\subseteq Y$, it follows $F=F\cap Y$, which says that $F$ is $\tau_{Y}$-closed by part (i).

To prove part (iii), since $\tau_{X}\mbox{-cl}(A)$ is $\tau_{X}$-closed in $X$, part (i) says that $Y\cap\tau_{X}\mbox{-cl}(A)$ is $\tau_{Y}$-closed in $Y$. Since $Y\cap\tau_{X}\mbox{-cl}(A)$ contains $A$, it follows
\[\tau_{Y}\mbox{-cl}(A)\subseteq Y\cap\tau_{X}\mbox{-cl}(A).\]
On the other hand, since $\tau_{Y}\mbox{-cl}(A)$ is $\tau_{Y}$-closed in $Y$, using part (i) again, there exists a $\tau_{X}$-closed set $F_{X}$ in $X$ satisfying
\[A\subseteq\tau_{Y}\mbox{-cl}(A)=Y\cap F_{X},\]
which implies $A\subseteq F_{X}$, i.e., $\tau_{X}\mbox{-cl}(A)\subseteq F_{X}$. Therefore, we obtain
\[Y\cap\tau_{X}\mbox{-cl}(A)\subseteq Y\cap F_{X}=\tau_{Y}\mbox{-cl}(A).\]
If $Y$ is $\tau_{X}$-closed, then $Y=\tau_{X}\mbox{-cl}(Y)$. Since $A\subseteq Y$, we have
\[\tau_{X}\mbox{-cl}(A)\subseteq\tau_{X}\mbox{-cl}(Y)=Y,\]
which shows
\[Y\cap\tau_{X}\mbox{-cl}(A)=\tau_{X}\mbox{-cl}(A).\]
This completes the proof. $\blacksquare$

\begin{equation}{\label{top75}}\tag{16}\mbox{}\end{equation}
Proposition \ref{top75}. Let $(Y,\tau_{Y})$ be a subspace of the topological space $(X,\tau_{X})$. Given any $x\in Y$, we have the following properties.

(i) Suppose that $N_{x}$ is $\tau_{X}$-neighborhood of $x$. Then $N_{x}\cap Y$ is also a $\tau_{Y}$-neighborhood of $x$.

(ii) Suppose that $Y$ is $\tau_{X}$-open, and that $N_{x}$ is $\tau_{Y}$-neighborhood of $x$. Then $N_{x}$ is also a $\tau_{X}$-neighborhood of $x$.

Proof. To prove part (i), there exists a $\tau_{X}$-open set $O$ satisfying $x\in O\subseteq N_{x}$. Therefore, we have
\[x\in O\cap Y\subseteq N_{x}\cap Y.\]
Since $O\cap Y$ is $\tau_{Y}$-open, it says that $N_{x}\cap Y$ is a $\tau_{Y}$-neighborhood of $x$. To prove part (ii), suppose that $N_{x}$ is $\tau_{Y}$-neighborhood of $x$. There exists a $\tau_{Y}$-open set $O$ satisfying $x\in O\subseteq N_{x}\subseteq Y$. Using part (ii) of Proposition \ref{top74}, we see that $O$ is also $\tau_{X}$-open, which shows that $N_{x}$ is a $\tau_{X}$-neighborhood of $x$. This completes the proof. $\blacksquare$

In general, we have
\[\tau_{Y}\mbox{-int}(A)\neq Y\cap\tau_{X}\mbox{-int}(A)\mbox{ and }\tau_{Y}\mbox{-bd}(A)\neq Y\cap\tau_{X}\mbox{-bd}(A).\]
The counterexamples are given below.

Example. Let $X=\mathbb{R}^{2}$ be endowed with the usual topology induced by the Euclidean metric

\[d\left (x_{1},y_{1}),(x_{2},y_{2})\right )=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}.\]
We consider the subset
\[Y=\{(x,y):y=0\},\]
i.e., the $x$-axis. If $A=Y$, then $\tau_{Y}\mbox{-int}(A)=Y$. However, since $Y$ contains no $\tau_{X}$-open subset in $X=\mathbb{R}^{2}$, it follows
\[Y\cap\tau_{X}\mbox{-int}(A)=\emptyset.\]
We also have
\[\tau_{Y}\mbox{-bd}(A)=\emptyset\mbox{ and }\tau_{X}\mbox{-bd}(A)=A,\]
which says
\[\tau_{Y}\mbox{-bd}(A)\neq Y\cap\tau_{X}\mbox{-bd}(A).\]

\begin{equation}{\label{f}}\tag{F}\mbox{}\end{equation}

Neighborhood System.

Definition. Let $(X,\tau )$ be a topological space. The neighborhood system ${\cal N}_{x}$ of a point $x\in X$ is the family of all neighborhoods of the point $x$. $\sharp$

\begin{equation}{\label{top8}}\tag{17}\mbox{}\end{equation}
Proposition \ref{top8}. Let $(X,\tau )$ be a topological space. Given $x\in X$, let ${\cal N}_{x}$ be a neighborhood system of $x$. Then, we have the following properties.

(a) If $N\in {\cal N}_{x}$, then $x\in N$.

(b) If $N_{1},N_{2}\in {\cal N}_{x}$, then $N_{1}\cap N_{2}\in {\cal N}_{x}$.

(d) If $N_{1}\in {\cal N}_{x}$, then there exists $N_{2}\in {\cal N}_{x}$ satisfying $N_{2}\subseteq N_{1}$ and $N_{2}\in {\cal N}_{y}$ for each $y\in N_{2}$ (i.e., $N_{2}$ is a neighborhood of each of its points).

Conversely, given a set $X$ and any $x\in X$, if the family ${\cal N}_{x}$ of subsets of $X$ satisfies the properties (a) and (b), then $X$ can be endowed with a topology $\tau$ such that $O\in\tau$ if and only if, for each $x\in O$, there exists $N\in {\cal N}_{x}$ satisfying $x\in N\subseteq O$. If properties (c) and (d) are satisfied, then ${\cal N}_{x}$ is a neighborhood system of $x$ with respect to this topology $\tau$.

Proof. Property (a) is obvious. Suppose that $N_{1}$ and $N_{2}$ are neighborhoods of $x$. Then, there exist open sets $O_{1}$ and $O_{2}$ contained in $N_{1}$ and $N_{2}$, respectively, which also says that $N_{1}\cap N_{2}$ contains the open set $O_{1}\cap O_{2}$. This shows that $N_{1}\cap N_{2}$ is a neighborhood of $x$, and shows property (b). Suppose that $N_{1}$ is a neighborhood of $x$. Then, there exists $O\in\tau$ satisfying $x\in O\subseteq N_{1}$. For $N_{1}\subseteq N_{2}$, we also have $x\in O\subseteq N_{2}$, i.e., $N_{2}$ is a neighborhood of $x$, which proves property (c). For property (d), we can take $N_{2}=\mbox{int}(N_{1})$. Then $N_{2}$ is an open set contained in $N_{1}$, which says that $N_{2}$ is a neighborhood of each of its points.

For the converse, we say that $O\in\tau$ if and only if, for each $x\in O$, there exists $N\in {\cal N}_{x}$ satisfying $N\subseteq O$. We consider axioms of topology.

It is clear to see $X\in\tau$ and $\emptyset\in\tau$.
Given any $O_{1},O_{2}\in\tau$ and any $x\in O_{1}\cap O_{2}$, there exists $N_{1},N_{2}\in {\cal N}_{x}$ satisfying $N_{1}\subseteq O_{1}$ and $N_{2}\subseteq O_{2}$. Property (b) says $N_{1}\cap N_{2}\in {\cal N}_{x}$ and $N_{1}\cap N_{2}\subseteq O_{1}\cap O_{2}$. This shows $O_{1}\cap O_{2}\in\tau$.
Let $\{O_{\gamma}\}_{\gamma\in\Gamma}$ be a family of members in $\tau$. For $x\in O=\bigcup_{\gamma\in\Gamma}O_{\gamma}$, there exists $O_{\gamma}$ satisfying $x\in O_{\gamma}$. This also says that there exists $N_{\gamma}\in {\cal N}_{x}$ satisfying $N_{\gamma}\subseteq O_{\gamma}\subseteq O$, which shows $O\in\tau$.

Therefore, we conclude that $(X,\tau)$ is a topological space. Finally, we want to show that ${\cal N}_{x}$ is a neighborhood system of $x$ when properties (c) and (d) are satisfied. Given any $N\in {\cal N}_{x}$, property (d) says that there exists $O\in {\cal N}_{x}$ satisfying $O\subseteq N$ and $O\in {\cal N}_{y}$ for each $y\in O$. This says $O\in\tau$. Therefore, every element in ${\cal N}_{x}$ is a neighborhood of $x$. Now, we need to claim that each neighborhood of $x$ is an element of ${\cal N}_{x}$. Let $N$ be a neighborhood of $x$. There exists $O\in\tau$ satisfying $x\in O\subseteq N$. The definition of $\tau$ says that there exists $N_{0}\in {\cal N}_{x}$ satisfying $N_{0}\subseteq O\subseteq N$. Using property (c), it follows $N\in {\cal N}_{x}$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top67}}\tag{18}\mbox{}\end{equation}
Proposition \ref{top67}. Let $\tau_{1}$ and $\tau_{2}$ be two topologies on the same set $X$. Let ${\cal N}_{x}^{(1)}$ and ${\cal N}_{x}^{(2)}$ be the neighborhood systems of $x$ for $\tau_{1}$ and $\tau_{2}$, respectively. Then $\tau_{1}\subseteq\tau_{2}$ if and only if, for each $x\in X$ and $N_{x}^{(1)}\in {\cal N}_{x}^{(1)}$, there exists $N_{x}^{(2)}\in {\cal N}_{x}^{(2)}$ satisfying $N_{x}^{(2)}\subseteq N_{x}^{(1)}$.

Proof. Let $O$ be $\tau_{1}$-open and $x\in O$. Then, there exists $N_{x}^{(1)}\in {\cal N}_{x}^{(1)}$ satisfying $x\in N_{x}^{(1)}\subseteq O$. If there exists $N_{x}^{(2)}\in {\cal N}_{x}^{(2)}$ satisfying $x\in N_{x}^{(2)}\subseteq N_{x}^{(1)}$, then $x\in N_{x}^{(2)}\subseteq O$, which shows that $O$ is $\tau_{2}$-open, i.e., $\tau_{1}\subseteq\tau_{2}$. On the other hand, suppose that $\tau_{1}\subseteq\tau_{2}$. If $N_{x}^{(1)}$ is a neighborhood of $x$ for $\tau_{1}$, then $N_{x}^{(1)}$ is also a neighborhood of $x$ for $\tau_{2}$. Using part (iv) of Proposition \ref{top8}, there exists $N_{x}^{(2)}\in {\cal N}_{x}^{(2)}$ satisfying $N_{x}^{(2)}\subseteq N_{x}^{(1)}$. This completes the proof. $\blacksquare$

Definition. Let ${\cal N}_{x}$ be a neighborhood system of $x$, and let ${\cal B}_{x}$ be a subfamily of ${\cal N}_{x}$. We say that ${\cal B}_{x}$ is a neighborhood base at $x$ when, given any $N\in {\cal N}_{x}$, there exists $B\in {\cal B}_{x}$ satisfying $B\subseteq N$. $\sharp$

It is clear to see that the family of all open sets containing $x$ is a neighborhood base at $x$.

Definition. A topological space is said to satisfy the first axiom of countability when every point has a countable neighborhood base. $\sharp$

Suppose that the family $\{N_{x}^{(n)}\}_{n=1}^{\infty}$ is a countable local base at $x$. Then, we can construct a new local base $\{M_{x}^{(n)}\}_{n=1}^{\infty}$ at $x$ satisfying $M_{x}^{(n+1)}\subseteq M_{x}^{(n)}$ for each $n$, where
\[M_{x}^{(n)}=\bigcap_{k=1}^{n}N_{x}^{(k)}.\]

Definition. Let ${\cal S}_{x}$ be a subfamily of the neighborhood system ${\cal N}_{x}$ of $x$. We say that ${\cal S}_{x}$ is a neighborhood subbase at $x$ when the family of finite intersections of members of ${\cal S}_{x}$ is a neighborhood base at $x$. $\sharp$

Suppose that the family $\{N_{x}^{(n)}\}_{n=1}^{\infty}$ is a countable local subbase at $x$. Then, the family $\{M_{x}^{(n)}\}_{n=1}^{\infty}$ defined by
\[M_{x}^{(n)}=\bigcap_{k=1}^{n}N_{x}^{(k)}\]
is a countable local base at $x$. This says that the existence of a countable local subbase at each point implies the first axiom of countability.

Definition. A topological space is said to satisfy the second axiom of countability when there exists a countable base for the topology. $\sharp$

Definition. Let $A$ be a subset of a topological space $(X,\tau )$.

We say that $A$ is {\bf dense} in $X$ when $\mbox{cl}(A)=X$.
The topological space $(X,\tau )$ is said to be separable when there exists a countable subset which is dense in $X$. $\sharp$

A separable space may fail to satisfy the second axiom of countability. However, the converse is true given below

\begin{equation}{\label{top45}}\tag{19}\mbox{}\end{equation}
Proposition \ref{top45}. A topological space that has a countable base is separable. Moreover, if we choose a point from each member of the base, then this countable set is a dense subset.

Proof. By choosing a point from each member of the base, we obtain a countable set $A$. We want to claim that the complement of the closure of $A$ is empty. Suppose that the complement set of the closure of $A$ contains some members of ${\cal B}$. Then, the complement set contains some points of $A$, which is a contradiction. Therefore, the complement of the closure of $A$ is an open set which contains no member of the base ${\cal B}$. Suppose that the complement set is nonempty. Since the complement set is also open, it is the union of some members of ${\cal B}$. Therefore, the contradiction occurs, since the complement set contains no members of ${\cal B}$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top46}}\tag{20}\mbox{}\end{equation}
Proposition \ref{top46}. Let $(X,\tau)$ be a topological space. A subset $D$ of $X$ is dense if and only if every open subset of $X$ contains some point of $D$. $\sharp$

Proposition. Any open subspace of a separable space is separable.

Proof. Suppose that $\{x_{n}\}_{n=1}^{\infty}$ is a countable dense subset of the separable space $(X,\tau_{X})$. Let $(Y,\tau_{Y})$ be a any $\tau_{X}$-open subspace of $X$, and let $D=\{x_{n}\}_{n=1}^{\infty}\cap Y$. Let $O$ be any $\tau_{Y}$-open subset of $Y$; that is, $O=Y\cap O^{*}$ for some $\tau_{X}$-open set $O^{*}$. Since $Y$ is $\tau_{X}$-open, it follows that $O$ is also $\tau_{X}$-open. Therefore $O$ contains some points $x_{n}$ by Proposition \ref{top46}, i.e., $O\cap D\neq\emptyset$. Using Proposition \ref{top46} again, it follows that $D$ is dense in $Y$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top47}}\tag{21}\mbox{}\end{equation}

Remark \ref{top47}.We have the following observations.

Each topological space which satisfies the second axiom of countability also satisfies the first axiom of countability.
Every metric space satisfies the first axiom of countability, since the open balls centered at $x$ with rational radii are countable and form a base at $x$.
Proposition \ref{top6} states that a metric space satisfies the second axiom of countability if and only if it is separable.
Proposition \ref{top45} shows that each topological space which satisfies the second axiom of countability is separable.
Any subspace of a topological space satisfying the first axiom of countability also satisfies the first axiom of countability.
Any subspace of a topological space satisfying the second axiom of countability also satisfies the second axiom of countability. $\sharp$

\begin{equation}{\label{g}}\tag{G}\mbox{}\end{equation}

Continuity.

Let $f$ be a function defined on $X$ into $Y$. Given a subset $A$ of $X$ and a subset $B$ of $Y$. The image $f(A)$ of $A$ is the set defined by
\[f(A)=\left\{f(x):x\in A\right\}.\]
The inverse image of $B$ under $f$ is defined by
\[f^{-1}(B)=\left\{x:f(x)\in B\right\}.\]
Let $\{B_{\gamma}\}_{\gamma\in\Gamma}$ be a family of subsets of $Y$. Then, we have
\[f^{-1}\left (\bigcap_{\gamma\in\Gamma}B_{\gamma}\right )=\bigcap_{\gamma\in\Gamma}f^{-1}\left (B_{\gamma}\right )\]

and

\[f^{-1}\left (\bigcup_{\gamma\in\Gamma}B_{\gamma}\right )=\bigcup_{\gamma\in\Gamma}f^{-1}\left (B_{\gamma}\right ).\]
Let $\{A_{\gamma}\}_{\gamma\in\Gamma}$ be a family of subsets of $X$. Then, we have
\[f\left (\bigcup_{\gamma\in\Gamma}A_{\gamma}\right )=\bigcup_{\gamma\in\Gamma}f\left (A_{\gamma}\right ).\]
However, we cannot have the similar equality regarding the intersection.

Definition. Let $f$ be a function from a topological space $(X,\tau_{X})$ into another topological space $(Y,\tau_{Y})$.

We say that $f$ is continuous when the inverse image of each $\tau_{Y}$-open set is $\tau_{X}$-open. More precisely, if $O\in\tau_{Y}$ then $f^{-1}(O)\in\tau_{X}$.
We say that $f$ is continuous at a point $x\in X$ when the inverse image of each $\tau_{Y}$-neighborhood of $f(x)$ is a $\tau_{X}$-neighborhood of $x$. $\sharp$

It is clear to see that $f$ is continuous if and only if it is continuous at each point of its domain.

Example. Suppose that $(X,\tau_{X})$ is a discrete topological space, and that $(Y,\tau_{Y})$ is any topological space. Then, any function $f$ from $(X,\tau_{X})$ into $(Y,\tau_{Y})$ is continuous. Indeed, given any subset $O_{Y}$ of $Y$, the inverse image $f^{-1}(O_{Y})$ is $\tau_{X}$-open
in $X$, since any subset of $X$ is $\tau_{X}$-open. $\sharp$

\begin{equation}{\label{top39}}\tag{22}\mbox{}\end{equation}
Proposition \ref{top39}. A function $f$ from a topological space $(X,\tau_{X})$ into a topological space $(Y,\tau_{Y})$ is continuous if and only if, given any $f(x)\in Y$ and any neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ satisfying $f(U)\subseteq V$.

Proof. Suppose that $f$ is continuous. Let $V$ be any neighborhood of $f(x)$. Then, there exists a $\tau_{Y}$-open set $O_{Y}$ in $Y$ satisfying $f(x)\in O_{Y}\subseteq V$. The continuity says that $f^{-1}(O_{Y})$ is $\tau_{X}$-open. We also have $x\in f^{-1}(O_{Y})$. Let $U=f^{-1}(O_{Y})$. Then $U$ is a neighborhood of $x$ satisfying $f(U)=O_{Y}\subseteq V$.

For the converse, given any $\tau_{Y}$-open set $O_{Y}$, we want to claim that $f^{-1}(O_{Y})$ is $\tau_{X}$-open. Given any $z\in f^{-1}(O_{Y})$, i.e., $f(z)\in O_{Y}$, it means that $O_{Y}$ is a neighborhood of $f(z)$. The assumption says that there exists a neighborhood $U$ of $z$ satisfying $f(U)\subseteq O_{Y}$, which also says that there exists a $\tau_{X}$-open set $O_{X}$ satisfying $z\in O_{X}\subseteq U$ and $f(U)\subseteq O_{Y}$. Therefore, we obtain
\[f(O_{X})\subseteq f(U)\subseteq O_{Y},\]
which implies $O_{X}\subseteq f^{-1}(O_{Y})$. This shows that, for each $z\in f^{-1}(O_{Y})$, there exists a $\tau_{X}$-open set $O_{X}$ satisfying $O_{X}\subseteq f^{-1}(O_{Y})$, which also says that $f^{-1}(O_{Y})$ is the union of $\tau_{X}$-open subsets of $X$. It follows that $f^{-1}(O_{Y})$ is $\tau_{X}$-open. This completes the proof. $\blacksquare$

\begin{equation}{\label{top16}}\tag{23}\mbox{}\end{equation}
Proposition \ref{top16}. Let $f$ be a function from the topological space $(X,\tau_{X})$ into the topological space $(Y,\tau_{Y})$. Let ${\cal B}_{Y}$ be any base for $\tau_{Y}$. Then $f$ is continuous if and only if, given any $B\in {\cal B}_{Y}$, the inverse image $f^{-1}(B)$ is $\tau_{X}$-open.

Proof. Suppose that $f$ is continuous. Since each $B\in {\cal B}_{Y}$ is $\tau_{Y}$-open, it follows that $f^{-1}(B)$ is $\tau_{X}$-open. For the converse, let $O_{Y}$ be any $\tau_{Y}$-open set. Proposition~\ref{top10} says that there exists an index set $\Gamma$ satisfying $O_{Y}=\bigcup_{\gamma\in\Gamma}B_{\gamma}$, where $B_{\gamma}\in {\cal B}_{Y}$ for all $\gamma\in\Gamma$. It follows
\[f^{-1}(O_{Y})=f^{-1}\left (\bigcup_{\gamma\in\Gamma}B_{\gamma}\right )=\bigcup_{\gamma\in\Gamma}f^{-1}(B_{\gamma}).\]
Since each $f^{-1}(B_{\gamma})$ is $\tau_{X}$-open, it says that $f^{-1}(O_{Y})$ is also $\tau_{X}$-open. This completes the proof. $\blacksquare$

\begin{equation}{\label{top65}}\tag{24}\mbox{}\end{equation}
Proposition \ref{top65}. A function $f$ from a topological space $(X,\tau_{X})$ into a topological space $(Y,\tau_{Y})$ is continuous if and only if the inverse image of every $\tau_{Y}$-closed set is $\tau_{X}$-closed. $\sharp$

Proposition. Let $X$ be any set endowed with two topologies $\tau_{1}$ and $\tau_{2}$. Then $\tau_{1}$ is finer than $\tau_{2}$ if and only if the identity function $i$ from $(X,\tau_{1})$ into $(X,\tau_{2})$ defined by $i(x)=x$ for all $x\in X$ is continuous. $\sharp$

Proposition. Suppose that $f$ is a continuous function from $(X,\tau_{X})$ into $(Y,\tau_{Y})$, and that $g$ is a continuous function from $(Y,\tau_{Y})$ into $(Z,\tau_{Z})$. Then, the composition $g\circ f$ is a continuous function from $(X,\tau_{X})$ into $(Z,\tau_{Z})$.

Proof. Suppose that $O_{Z}$ is $\tau_{Z}$-open. Then $g^{-1}(O_{Z})$ is $\tau_{Y}$-open. Since
\[(g\circ f)^{-1}(O_{Z})=f^{-1}(g^{-1}(O_{Z}))\]
is $\tau_{X}$-open, it says that $g\circ f$ is continuous. This completes the proof. $\blacksquare$

Proposition. Suppose that $f$ is a continuous function from $(X,\tau_{X})$ into $(Z,\tau_{Z})$. Then, the restriction $f|_{Y}$ of $f$ to a subspace $(Y,\tau_{Y})$ of $(X,\tau_{X})$ is a continuous function from $(Y,\tau_{Y})$ into $(Z,\tau_{Z})$.

Proof. Give any $\tau_{Z}$-open set $O_{Z}$, the inverse image $f^{-1}(O_{Z})$ is $\tau_{X}$-open. Since $f|_{Y}^{-1}(O_{Z})=Y\cap f^{-1}(O_{Z})$ is $\tau_{Y}$-open for any $\tau_{Z}$-open set $O_{Z}$, this completes the proof. $\blacksquare$

Suppose that $X$ is any set, and that $(Y,\tau_{Y})$ is a topological space. Let $f$ be a function from $X$ into $Y$. We expect to find a topology on $X$ such that $f$ is continuous. It is obvious that the discrete and trivial topology can achieve this expectation. Define the following family
\[\tau_{X}=\left\{f^{-1}(O_{Y}):O_{Y}\in\tau_{Y}\right\}.\]
We can show that $(X,\tau_{X})$ is the coarsest topology such that $f$ is continuous.

Definition. A homeomorphism between two topological spaces is a one-to-one continuous function of $(X,\tau_{X})$ onto $(Y,\tau_{Y})$ for which $f^{-1}$ is also continuous. The topological spaces $(X,\tau_{X})$ and $(Y,\tau_{Y})$ are said to be homeomorphic when there exists a homeomorphism between them. Two topological spaces are topologically equivalent when they are homeomorphic. A function $g$ from $(X,\tau_{X})$ into $(Y,\tau_{Y})$ is said to be an embedding of $X$ into $Y$ when $g$ is a homeomorphism from $X$ onto a subspace of $Y$. $\sharp$

Example. Let $\mathbb{R}^{2}$ be endowed with the usual topology. Then, any two closed line segments in $\mathbb{R}^{2}$ are homeomorphic, and
any triangle is homeomorphic to any circle. $\sharp$

Proposition. Let $f$ be a one-to-one function from $(X,\tau_{X})$ onto $(Y,\tau_{Y})$. Then, the following statements are equivalent.

(a $f$ is a homeomorphism.

(b) A subset $O_{Y}$ of $Y$ is $\tau_{Y}$-open if and only if $f^{-1}(O_{Y})$ is $\tau_{X}$-open.

(d) If ${\cal B}_{X}$ is any base for $\tau_{X}$, then $\{f(B):B\in {\cal B}_{X}\}$ is a base for $\tau_{Y}$. $\sharp$

The identity function of a topological space onto itself is always a homeomorphism, and the inverse of a homeomorphism is again a homeomorphism. It is also evident that the composition of two homeomorphism is a homeomorphism. Consequently, the collection of topological spaces can be divided into equivalence classes such that each topological space is homeomorphic to every member of its equivalence class and to these spaces only. We have the following observations.

Two discrete spaces, $X$ and $Y$, are homeomorphic if and only if there is a one-to-one function from $X$ onto $Y$; that is, if and only if $X$ and $Y$ have the same cardinal number. This is true because every function on a discrete space is continuous, regardless of the topology of the range space. It is also true that two indiscrete spaces (the only open sets are the space and the empty set) are homeomorphic if and only if there is a one-to-one map of one onto the other, because each function into indiscrete space is continuous regardless of the topology of the domain space.
The set of all real numbers with the usual topology is homeomorphic to the open interval $(0,1)$ with the relative topology, since the function whose value at a member $x$ of $(0,1)$ is $(2x-1)/x(x-1)$ is easily proved to be a homeomorphism. However, the interval $(0,1)$ is not homeomorphic to $(0,1)\cup (1,2)$, This little demonstration was achieved by noticing that one of the space is connected, the other is not .

\begin{equation}{\label{top218}}\tag{25}\mbox{}\end{equation}
Proposition \ref{top218}. Let $f$ be a function from $(X,\tau_{X})$ onto $(Y,\tau_{Y})$. Then, the following statements are equivalent.

(a) The function $f$ is continuous.

(b) The inverse image of each $\tau_{Y}$-closed set is $\tau_{X}$-closed.

(c) For each $x$ in $X$, the inverse image of every $\tau_{Y}$-neighborhood of $f(x)$ is a $\tau_{X}$-neighborhood of $x$.

(e) For each $x$ in $X$ and each $\tau_{Y}$-neighborhood $U$ of $f(x)$, there exists a $\tau_{X}$-neighborhood $V$ of $x$ satisfying $f(V)\subseteq U$.

(f) Let $\{x_{\gamma}\}_{\gamma\in\Gamma}$ be a net in $X$ converging to $x\in X$. Then, the net $\{f(x_{n})\}_{\gamma\in\Gamma}$ converges to $f(x)$.

(g) For each subset $A$ of $X$, we have
\[f(\tau_{X}\mbox{-}\mbox{\em cl}(A))\subseteq\tau_{Y}\mbox{-}\mbox{\em cl}(f(A)).\]

(h) For each subset $B$ of $Y$, we have
\[\tau_{Y}\mbox{-}\mbox{\em cl}(f^{-1}(B))=f^{-1}(\tau_{X}\mbox{-}\mbox{\em cl}(B)).\]

\begin{equation}{\label{h}}\tag{H}\mbox{}\end{equation}

Quotient Space and Quotient Topology.

Let $(X,\tau)$ be a topological space, and let “$\sim$” be an equivalence relation on $X$. Let $[X]$ denote the collection of all equivalence classes $[x]$ for $x\in X$. The function $f:X\rightarrow [X]$ defined by $f(x)=[x]$ is called an quotient function.

Proposition. Let $f$ be a quotient function from $(X,\tau_{X})$ into $[X]$. Define the family
\[\tau_{[X]}=\left\{O_{[X]}\subseteq [X]:f^{-1}\left (O_{[X]}\right )\in\tau_{X}\right\}.\]
Then $([X],\tau_{[X]})$ forms a topological space.

Proof. Since $f^{-1}(\emptyset )=\emptyset$ and $f^{-1}([X])=X$, it follows that $\emptyset$ and $[X]$ are in $\tau_{[X]}$. Given any $O_{[X]}^{(1)},O_{[X]}^{(2)}\in\tau_{[X]}$, it means $f^{-1}(O_{[X]}^{(1)}),f^{-1}(O_{[X]}^{(2)})\in\tau_{X}$. Since
\[f^{-1}\left (O_{[X]}^{(1)}\cap O_{[X]}^{(2)}\right )=f^{-1}\left (O_{[X]}^{(1)}\right )\capf^{-1}\left (O_{[X]}^{(2)}\right )\in\tau_{X},\]
this says $O_{[X]}^{(1)}\cap O_{[X]}^{(2)}\in\tau_{[X]}$. Now, let $\{O_{[X]}^{(\alpha )}\}_{\alpha\in\Lambda}$ be a family of members of $[X]$.
Then, we have $f^{-1}(O_{[X]}^{(\alpha )})\in\tau_{X}$-open. Since
\[f^{-1}\left (\bigcup_{\alpha\in\Lambda}O_{[X]}^{(\alpha )}\right )=\bigcup_{\alpha\in\Lambda}f^{-1}\left (O_{[X]}^{(\alpha )}\right )\in\tau_{X},\]
this says $\bigcup_{\alpha\in\Lambda}O_{[X]}^{(\alpha )}\in\tau_{[X]}$. This completes the proof. $\blacksquare$

The above proposition suggests the following definition.

Definition. Let $f$ be a quotient function from $(X,\tau_{X})$ into $[X]$. The quotient topology for $[X]$ is defined by the family

\[\tau_{[X]}=\left\{O_{[X]}\subseteq [X]:f^{-1}\left (O_{[X]}\right )\in\tau_{X}\right\}.\]
Also, $([X],\tau_{[X]})$ is called a quotient topological space. $\sharp$

Proposition. The quotient function $f$ from the topological space $(X,\tau_{X})$ into the quotient topological space $([X],\tau_{[X]})$ is continuous. Moreover, the quotient topology $\tau_{[X]}$ is the finest topology for $[X]$ such that $f$ is continuous.

Proof. It is obvious that the quotient function $f$ defined in this way is continuous. Let $\tau_{[X]}^{*}$ be another topology for $[X]$ such that it is strictly finer than $\tau_{[X]}$. Then, there is $O\in\tau_{[X]}^{*}$ satisfying $O\not\in\tau_{[X]}$, which says $f^{-1}(O)\not\in\tau_{X}$. It also means that $f$ is not continuous from $X$ into $[X]$. This completes the proof. $\blacksquare$

Example. Assume that the closed interval $[0,1]$ is endowed with the usual topology. Let $0$ be equivalent to $1$, and let every other element of $(0,1)$ be equivalent only to itself. In this case, the equivalent classes are $\{0,1\}$ and $\{x\}$ for $x\in (0,1)$. By identifying $0$ and $1$, we obtain a circle, which represents the quotient topological space. Therefore, we have a continuous function from $[0,1]$ onto a circle. $\sharp$

Let $g$ be a function from $X$ into $Y$, and let “$\sim$” be an equivalent relation on $X$ defined by
\[x_{1}\sim x_{2}\mbox{ if and only if }g(x_{1})=g(x_{2})\mbox{ for all }x_{1},x_{2}\in X.\]
Let $[x]$ be an equivalent class. Then, there is a natural function $\hat{g}$ from $[X]$ into $Y$ defined by $\hat{g}([x])=g(x)$. The function $\hat{g}$ is well-defined, since $[x_{1}]=[x_{2}]$ implies
\[\hat{g}([x_{1}])=g(x_{1})=g(x_{2})=\hat{g}([x_{2}]).\]
Let $f$ be a quotient function from $X$ into $[X]$. Then, we have the following interesting result.

Proposition. Suppose that $g$ is onto, and that $[X]$ is endowed with the quotient topology $\tau_{[X]}$. Then $\hat{g}$ is a homeomorphism from $([X],\tau_{[X]})$ onto $(Y,\tau_{Y})$.

Proof. Since $g$ is onto, it follows that $\hat{g}$ is also onto. Suppose that $\hat{g}([x_{1}])=\hat{g}([x_{2}])$. Then $g(x_{1})=g(x_{2})$, which says $x_{1}\sim x_{2}$. Therefore, we have $[x_{1}]=[x_{2}]$, which shows that $\hat{g}$ is one-to-one. We remain to show that $\hat{g}$ and $\hat{g}^{-1}$ are continuous. Given any $O_{Y}\in\tau_{Y}$, since
\[f^{-1}(\hat{g}^{-1}(O_{Y})=g^{-1}(O_{Y})\in\tau_{X},\]
it follows $\hat{g}^{-1}(O_{Y})\in\tau_{[X]}$, which also shows that $\hat{g}$ is continuous. Now, given any $O_{[X]}\in\tau_{[X]}$, we have $f^{-1}(O_{[X]})\in\tau_{X}$, which says $f^{-1}(O_{[X]})=g^{-1}(O_{Y})$ for some $O_{Y}\in\tau_{Y}$. Since
\[(\hat{g}^{-1})^{-1}(O_{[X]})=\hat{g}(O_{[X]})=g(f^{-1}(O_{[X]})=g(g^{-1}(O_{Y}))=O_{Y}\in\tau_{Y},\]
it says that $\hat{g}^{-1}$ is continuous. This completes the proof. $\blacksquare$

Let $f$ be a function from a set $X$ into a topological space $Y$. Then, it is always possible to assign a topology to $X$ such that $f$ is a continuous function. An interesting topology is the family $\tau$ of all sets of the form $f^{-1}(U)$ for $U$ open in $Y$. This family is evidently a topology, since the inverse of a function preserves unions. We can show that $\tau$ is the coarsest topology such that $f$ is continuous.

Let $f$ be a function from a topological space $X$ into $Y$. We can also assign a topology to $Y$ such that $f$ is continuous. Let $U$ be a subset of $Y$ such that it is open in a topology relative to which $f$ is continuous. Then $f^{-1}(U)$ is open in $X$. On the other hand, the family $\eta$ of all subsets $U$ such that $f^{-1}(U)$ is open in $X$ is a topology for $Y$ since the inverse of an intersection (or union) of members of the family is the intersection (union) of the inverses. Therefore, this topology $\eta$ is the finest topology for $Y$ such that the function $f$ is continuous. It is also called the quotient topology for $Y$ (the quotient topology relative to $f$ and the topology of $X$). A subset $B$ of $Y$ is closed relative to the quotient topology if and only if

\[f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)\]

is open in $X$. In other words, $B$ is closed if and only if $f^{-1}(B)$ is closed.

Definition. A function $f$ from a topological space into another topological space is said to be open when the image of each open set is open. A function $f$ is said to be closed when the image of each closed set is closed.

Proposition. Suppose that $f$ is a continuous function from the topological space $(X,\tau )$ onto the topological space $(Y,\eta )$ such that $f$ is either open or closed. Then $\eta$ is the quotient topology. $\sharp$

Proposition. Let $f$ be a function from a topological space into a product space. Then $f$ is continuous if and only if the composition of $f$ with each projection is continuous.

Proposition. Let $f$ be a continuous function of a space $X$ onto a space $Y$, and let $Y$ have the quotient topology. Then a function $g$ on $Y$ to a topological space $Z$ is continuous if and only if the composition $g\circ f$ is continuous. $\sharp$

Definition. A decomposition (partition) of $X$ is a disjoint family ${\cal D}$ of subsets of $X$ whose union is $X$. The projection (quotient map) of $X$ onto the decomposition ${\cal D}$ is the function $P$ whose value at $x$ is the unique member of ${\cal D}$ to which $x$ belongs. $\sharp$

There is an equivalent way of describing a decomposition. Given ${\cal D}$, define a relation $R$ on $X$ by agreeing that a point $x$ is $R$ related to a point $y$ if and only if $x$ and $y$ belong to the same member of the decomposition. Formally, the relation $R$ of the decomposition ${\cal D}$ is the subset of $X\times X$ consisting of all pairs $(x,y)$ such that $x$ and $y$ belong to the same member of ${\cal D}$,or, briefly,

\[R=\bigcup_{D\in {\cal D}}D\times D.\]
If $P$ is the projection of $X$ onto ${\cal D}$, then $R=\{(x,y):P(x)=P(y)\}$. The relation $R$ is an equivalence relation; that is, it is reflexive, symmetric, and transitive. Each equivalence relation on $X$ defines a family of subsets (the equivalence classes) which is a decomposition of $X$. If $R$ is an equivalence relation on $X$, then $X/R$ is defined to be the family of equivalence classes. If $A$ is a subset of $X$, then $R(A)$ is the set of all points which are $R$ relatives of points of $A$; that is,

\[R(A)=\{y:(x,y)\in R\mbox{ for some $x$ in $A$}\}.\]

Equivalently,

\[R(A)=\bigcup\{D:D\in X/R\mbox{ and }D\cap A\mbox{ is noempty }\}.\]

The set $R(x)$ is the equivalent class to which $x$ belongs, and if $P$ is the projection of $X$ onto the decomposition, then $P(x)=R(x)$.

Let $X$ be a fixed topological space, $R$ be an equivalence relation on $X$, and $P$ be the projection of $X$ onto the family $X/R$ of equivalence classes. The quotient space is the family $X/R$ with the quotient topology relative to $P$. If ${\cal A}\subseteq X/R$, then

\[P^{-1}({\cal A})=\bigcup_{A\in {\cal A}}P^{-1}(A)\]
and hence ${\cal A}$ is open (closed) relative to the quotient topology if and only if $\bigcup_{A\in {\cal A}}A$ is open (closed) in $X$.

Proposition. Let $P$ be the projection of the topological space $X$ onto the quotient space $X/R$. Then, the following statements are equivalent.

(a) $P$ is an open mapping.

(b) If $A$ is an open subset of $A$, then $R(A)$ is open.

If “open” and “closed” are interchanged in (a), (b), and (c) the resulting statements are still equivalent. $\sharp$

It is sometimes assumed that $R$, which is a set of ordered pairs, is closed in the product space. This condition may be restated: if $x$ and $y$ are members of $X$ which are not $R$ related, then there is a neighborhood $W$ of $(x,y)$ in the product space $X\times X$ which is disjoint from $R$. Such a neighborhood $W$ contains a neighborhood of the form $U\times V$, where $U$ and $V$ are neighborhoods of $x$ and $y$ respectively, and $U\times V$ is disjoint from $R$ if and only if there is no point of $U$ which is $R$ related to a point of $V$. That is, $R$ is closed in $X\times X$ if and only if, whenever $x$ and $y$ are points of $X$ which are not $R$ related, then there are neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that no point of $U$ is $R$ related to a point of $V$. Equivalently, no member of $X/R$ intersects both $U$ and $V$.

Proposition. If the quotient space $X/R$ is Hausdorff, then $R$ is closed in the product space $X\times X$. If the projection $P$ of a space $X$ onto the quotient space $X/R$ is open, and $R$ is closed in $X\times $latex X, then $X/R$ is a Hausdorff space. $\sharp$

A decomposition ${\cal D}$ of a topological space $X$ is upper semicontinuous when, for each $D$ in ${\cal D}$ and each open set $U$ containing $D$ there is an open set $V$ such that $D\subseteq V\subseteq U$ and $V$ is the union of members of ${\cal D}$.

Proposition. A decomposition ${\cal D}$ of a topological space $X$ is upper semicontinuous if and only if the projection $P$ of $X$ onto ${\cal D}$
is closed. $\sharp$

It is very easy to verify that ${\cal D}$ is upper semicontinuous, and if $X$ is Hausdorff, the relation $R=\bigcup\{D\times D:D\in {\cal D}\}$ is closed in $X\times X$.

\begin{equation}{\label{i}}\tag{I}\mbox{}\end{equation}

Connected Sets.

Definition. A topological space $X$ is said to be disconnected when $X$ can be expressed as the union of two nonempty disjoint open subsets of $X$. Otherwise, $X$ is said to be connected. A subset $A$ of $X$ is said to be connected when the subspace $A$ is connected. $\sharp$

Equivalently, a topological space is connected if and only if there do not exist two nonempty disjoint open sets $O_{1}$ and $O_{2}$ in $X$ satisfying $X=O_{1}\cup O_{2}$.

\begin{equation}{\label{top86}}\tag{26}\mbox{}\end{equation}
Remark \ref{top86}. We have the following observations

Suppose that the topological space $X$ is disconnected. Then, we have $X=O_{1}\cup O_{2}$, where $O_{1}$ and $O_{2}$ are disjoint open subset of $X$.
Since each set is the complement of the other, it follows that $O_{1}$ and $O_{2}$ are both closed and open in $X$.
A topological space $X$ is connected if and only if the only subsets of $X$ which are both open and closed are the sets $\emptyset$ and $X$.
A subset $A$ of a topological space is connected if and only if there do not exists open sets $O_{1}$ and $O_{2}$ in $X$ both meeting $A$ satisfying
\[A\subseteq O_{1}\cup O_{2}\mbox{ and }A\cap O_{1}\cap O_{2}=\emptyset.\]

\begin{equation}{\label{top91}}\tag{27}\mbox{}\end{equation}
Proposition \ref{top91}. Let $A$ be a subset of a topological space $(X,\tau )$. Then $A$ is connected if and only if $A$ cannot be expressed as $S\cup T$ satisfying
\begin{equation}{\label{top87}}\tag{28}
\mbox{cl}(S)\cap T=S\cap\mbox{cl}(T)=\emptyset ,
\end{equation}
where $S$ and $T$ are subsets of $X$.

Proof. We consider the subspace $(A,\tau_{A})$. Suppose that $A=S\cup T$ such that (\ref{top87}) is satisfied. Then, using Proposition \ref{top13},
\[\tau_{A}\mbox{-cl}(S)=A\cap\tau\mbox{-cl}(S)=(S\cup T)\cap\tau\mbox{-cl}(S)
=(S\cap\tau\mbox{-cl}(S))\cup (T\cap\tau\mbox{-cl}(S))=S\cup\emptyset =S,\]
which says that $S$ is $\tau_{A}$-closed. We can similarly show that $T$ is also $\tau_{A}$-closed. In this case, we also see that $S$ and $T$ are also $\tau_{A}$-open. The second observation of Remark \ref{top86} says that $A$ is disconnected.

On the other hand, suppose that $A$ is disconnected. Remark \ref{top86} says that $A=S\cup T$ can be expressed as the union of nonempty disjoint sets $S$ and $T$ such that $S$ and $T$ are both $\tau_{A}$-open and $\tau_{A}$-closed. Given $x\in S\cap\tau\mbox{-cl}(T)$, since $S\subseteq A$, Proposition \ref{top13} says
\[x\in A\cap\tau\mbox{-cl}(T)=\tau_{A}\mbox{-cl}(T)=T.\]
Therefore, we obtain $x\in S\cap T$, which contradicts $S$ and $T$ are disjoint. Therefore, we must have $S\cap\mbox{cl}(T)=\emptyset$. We can similarly show $\mbox{cl}(S)\cap T=\emptyset$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top93}}\tag{29}\mbox{}\end{equation}
Corollary \ref{top93}. Let $X$ be a topological space. We have the following properties

(i) If $X=S\cup T$ can be expressed as the union of nonempty disjoint open sets $S$ and $T$, then
\begin{equation}{\label{top92}}\tag{30}
\mbox{\em cl}(S)\cap T=S\cap\mbox{\em cl}(T)=\emptyset .
\end{equation}

(ii) If $X=S\cup T$ such that $(\ref{top92})$ is satisfied, then $S$ and $T$ are nonempty disjoint open subsets of $X$.

Proof. The desired results can be realized from the proof of Proposition \ref{top91} immediately. $\blacksquare$

Proposition. Let $A$ be a nonempty and proper subset of a topological space $(X,\tau )$. Then $A$ is connected if and only if $\mbox{bd}(A)\neq\emptyset$.

Proof. Suppose that $A$ is connected, and the $\mbox{bd}(A)=\emptyset$. Then, we have
\[\mbox{cl}(A)=\mbox{int}(A)\cup\mbox{bd}(A)=\mbox{int}(A),\]
which says that $A$ is both open and closed subset of $X$. Since $A$ is nonempty and $A\neq X$, the second observation of Remark \ref{top86} says that $A$ is disconnected. This contradiction says that$\mbox{bd}(A)\neq\emptyset$. For the converse, suppose that $A$ is disconnected. Remark \ref{top86}
says that $A=S\cup T$ can be expressed as the union of nonempty disjoint sets $S$ and $T$ such that $S$ and $T$ are both $\tau_{A}$-open and $\tau_{A}$-closed, i.e., $A=\mbox{cl}(A)=\mbox{int}(A)$, which says
\[\mbox{bd}(A)=\mbox{cl}(A)\setminus\mbox{int}(A)=A\setminus A=\emptyset .\]
This contradiction completes the proof. $\blacksquare$

\begin{equation}{\label{top88}}\tag{31}\mbox{}\end{equation}
Proposition \ref{top88}. Let $X$ be a topological space such that $X=S\cup T$ can be expressed as the union of nonempty disjoint open sets $S$ and $T$. If $A$ is any connected subset of $X$, then either $A\subseteq S$ or $A\subseteq T$.

Proof. By definition, the subspace $(A,\tau_{A})$ is a connected space. Suppose that $A\cap S\neq\emptyset$ and $A\cap T\neq\emptyset$. Then $A\cap S$ and $A\cap T$ are nonempty disjoint $\tau_{A}$-open sets such that $A=(A\cap S)\cup (A\cap T)$, which says that $A$ is disconnected. Therefore, we must have either $A\cap S=\emptyset$ or $A\cap T=\emptyset$. This completes the proof. $\sharp$

\begin{equation}{\label{top94}}\tag{32}\mbox{}\end{equation}
Proposition \ref{top94}. Suppose that $A$ is a connected subset of a topological space $X$. If $A\subseteq B\subseteq\mbox{cl}(A)$, then $B$ is also a connected subset of $X$. In particular, $\mbox{cl}(A)$ is also connected.

Proof. Suppose that $B$ is disconnected. Then $B=S\cup T$ can be expressed as the union of nonempty disjoint open sets $S$ and $T$. Since $A\subseteq B$, Proposition \ref{top88} says $A\subseteq S$ or $A\subseteq T$. If $A\subseteq S$, i.e., $\mbox{cl}(A)\subset\mbox{cl}(S)$, then $B\subseteq\mbox{cl}(A)\subseteq\mbox{cl}(S)$. Since $B=S\cup T$, it follows $T\subset\mbox{cl}(S)$, which contradicts $T\cap\mbox{cl}(S)=\emptyset$ by Corollary \ref{top93}. This completes the proof. $\blacksquare$

Corollary. If a topological space $X$ contains a connected dense subset, then $X$ is connected.

Proof. Let $A$ be the connected dense subset of $X$. Then $\mbox{cl}(A)=X$. The desired result follows immediately from Proposition \ref{top94}. $\blacksquare$

Proposition. The closure of a connected set is connected.

Proof. Suppose that $\mbox{cl}(A)$ is disconnected. Then, by Remark \ref{top86}, $\mbox{cl}(A)=S\cup T$, where $S$ and $T$ are nonempty disjoint subsets of $\mbox{cl}(A)$ and are both $\tau_{\mbox{cl}(A)}$-open and $\tau_{\mbox{cl}(A)}$-closed. Since $A\subset\mbox{cl}(A)$, it follows that $A\cap S$ and $A\cap T$ are both $\tau_{A}$-open and $\tau_{A}$-closed. Since $A=(A\cap S)\cup (A\cap T)$, it says that $A$ is disconnected. This contradiction completes the proof. $\blacksquare$

Proposition. Let $X$ be a topological space such that $X=\bigcup_{\alpha\in\Lambda}A_{\alpha}$, where each $A_{\alpha}$ is a connected subset of $X$. If $\bigcap_{\alpha\in\Lambda}A_{\alpha}\neq\emptyset$, then $X$ is itself connected.

Proof. Suppose that $X$ is disconnected. It means that $X=S\cup T$ can be expressed as the union of nonempty disjoint open sets $S$ and $T$. For each $\alpha\in\Lambda$, Proposition \ref{top88} says that we have either $A_{\alpha}\subseteq S$ or $A_{\alpha}\subseteq T$. Suppose that some $A_{\alpha}\subseteq S$. Since $\bigcap_{\alpha\in\Lambda}A_{\alpha}\neq\emptyset$ and $S\cap T=\emptyset$, we must have $A_{\alpha}\subseteq S$ for all $\alpha\in\Lambda$, which says $X=\bigcup_{\alpha\in\Lambda}A_{\alpha}\subseteq S$. It says $X=S$ and $T=\emptyset$, which contradicts $T\neq\emptyset$. Similarly, if some $A_{\alpha}\subseteq T$, then we can obtain $X=T$ and $S=\emptyset$. This contradiction completes the proof. $\blacksquare$

Proposition. Let $f$ be a continuous real-valued function defined on a connected space $X$. Given any $x,y\in X$ and $c\in\mathbb{R}$ such that $f(x)<c<f(y)$, there exists $z\in X$ satisfying $f(z)=c$.

Proof. Suppose that $f$ cannot have the value $c$. Then $f^{-1}((-\infty ,c))$ and $f^{-1}((c,\infty ))$ are nonempty disjoint open set such that their union is $X$. This shows that $X$ is disconnected, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top89}}\tag{33}\mbox{}\end{equation}
Proposition \ref{top89}. Let $f$ be a continuous function from a topological space $(X,\tau_{X})$ onto another topological space $(Y,\tau_{Y})$. If $X$ is connected, then $Y$ is also connected.

Proof. Suppose that $Y$ is disconnected. Then $Y=S\cup T$ can be expressed as the union of nonempty disjoint $\tau_{Y}$-open sets $S$ and $T$. Since $f$ is onto, it follows that $f^{-1}(S)$ and $f^{-1}(T)$ are nonempty disjoint subset of $X$ such that $X=f^{-1}(S)\cup f^{-1}(T)$. The continuity of $f$ says that $f^{-1}(S)$ and $f^{-1}(T)$ are $\tau_{X}$-open subsets of $X$. The second observation of Remark \ref{top86} says that $X$ is disconnected. This contradiction completes the proof. $\blacksquare$

\begin{equation}{\label{top90}}\tag{34}\mbox{}\end{equation}
Proposition \ref{top90}. Let $X$ be a topological space such that any two elements $x$ and $y$ in $X$ are contained in some connected subset of $X$. Then $X$ is connected.

Proof. Let $x$ be a fixed element in $X$. For any $y\in X$, let $C(x,y)$ be a connected subset of $X$ containing $x$ and $y$. Then $\bigcup_{y\in X}C(x,y)=X$ and $\bigcap_{y\in Y}C(x,y)\neq\emptyset$ since $x\in C(x,y)$ for each $y\in X$. Proposition \ref{top89} says that $X$ is connected, and the proof is complete. $\blacksquare$

Definition. Let $X$ be a topological space. A subspace $Y$ of $X$ is said to be a path when there exists a continuous function from $[0,1]$ that is endowed with the usual topology onto $Y$. The topological space $X$ is said to be path connected when, given any two points $x$ and $y$ in $X$, there exists a path in $X$ containing $x$ and $y$. $\sharp$

Proposition. If the topological space $X$ is path connected, then it is also connected.

Proof. Let $x$ be a fixed element in $X$. For any $y\in X$, let $P(x,y)$ denote a path containing $x$ and $y$. Then $P(x,y)$ is connected. We also have $\bigcup_{y\in X}P(x,y)=X$ and $\bigcap_{y\in Y}P(x,y)\neq\emptyset$, since $x\in P(x,y)$ for each $y\in X$. Proposition \ref{top89} says that $X$ is connected, and the proof is complete. $\blacksquare$

Definition. A subset $A$ of $\mathbb{R}^{m}$ is said to be polygonally connected when, given any two points $x$ and $y$ in $X$, there are points $\{x_{0}=x,x_{1},\cdots ,x_{n-1},x_{n}=y\}$ satisfying $\bigcup_{i=1}^{n}\overline{x_{i-1}x_{i}}\subseteq A$, where $\overline{x_{i-1}x_{i}}$ denotes the closed segment joining $x_{i-1}$ and $x_{i}$. $\sharp$

Proposition. Let $O$ be a connected and open subset of $\mathbb{R}^{n}$. Then $O$ is polygonally connected.

Proof. Given a fixed element $u\in O$, we define the set
\[A=\left\{a\in O:\mbox{$a$ can be polygonally connected to $u$}\right\}\mbox{ and }B=O\setminus A.\]
Then, we want to claim that $A$ is open. Given any $a\in A\subseteq O$, since $O$ is open, there exists a neighborhood $N_{a}$ of $a$ satisfying $N_{a}\subseteq O$. Since $N_{a}$ is convex, if $a$ can be polygonally connected to $a\in O$, then $u$ can also be polygonally connected in $O$ to any point of $N_{a}$, which says that $N_{a}\subseteq A$. This shows that $A$ is open. Since $B$ is the set of points in $O$ which cannot be polygonally connected to $u\in O$, a similar argument can show that $B$ is also open. In this case, we see that $O=A\cup B$, where $A$ and $B$ are disjoint union of open subsets of $O$. Since $O$ is connected, it follows that $A=\emptyset$ or $B=\emptyset$. Since $u\in A$, i.e., $A\neq\emptyset$, we must have $B=\emptyset$, which says $O=A$. This shows that $O$ is polygonally connected, and the proof is complete. $\blacksquare$

Definition. Let $X$ be a topological space. A maximal connected subset of $X$ is said to be a component of $X$. $\sharp$

Proposition. A component of a topological space is closed.

Proof. Let $A$ be the component. Then $A$ is connected. Proposition \ref{top94} says that $\mbox{cl}(A)$ is also connected. Since $A\subset\mbox{cl}(A)$ and $A$ is the maximal connected subset of $X$, it follows $A=\mbox{cl}(A)$. This completes the proof. $\blacksquare$

Proposition. Let $\{A_{\alpha}\}_{\alpha\in\Lambda}$ be a family of all components of a topological space. Then $X=\bigcup_{\alpha\in\Lambda}A_{\alpha}$ and $A_{\alpha}\cap A_{\beta}=\emptyset$ for $\alpha\neq\beta$.

Proof. For each $x\in X$, there exists a connected set containing $x$. Therefore, there exists a component containing $x$. This shows $X\subseteq\bigcup_{\alpha\in\Lambda}A_{\alpha}$. If $A_{\alpha}\cap A_{\beta}\neq\emptyset$, then $A_{\alpha}\cup A_{\beta}$ is a
connected subset of $X$, which contradicts the maximality. This completes the proof. $\blacksquare$

Definition. A topological space $X$ is said to be locally connected when, for any $x\in X$ and any neighborhood $N_{x}$ of $x$, there exists a connected neighborhood $U_{x}$ of $x$ satisfying $U_{x}\subseteq N_{x}$. $\sharp$

Proposition. Let $X$ be a topological space. Then, the following statements are equivalent.

(a) $X$ is locally connected.

(b) Each component of each open subspace of $X$ is open.

Corollary. A compact and locally connected space has at most finitely many components. $\sharp$

Proposition. If $f$ is a continuous open mapping from a locally connected space $(X,\tau_{X})$ onto a space $(Y,\tau_{Y})$, then $Y$ is also locally connected. $\sharp$

Proposition. The product space $\prod_{i=1}^{\infty}X_{i}$ of the countable family of nonempty topological spaces $\{(X_{i},\tau_{i})\}_{i=1}^{\infty}$ is locally connected if and only if each component space $X_{i}$ is locally connected and all but finitely many of the $X_{i}$ are connected. $\sharp$

Two subsets $A$ and $B$ are said to be separated in a topological space $X$ when
\[\mbox{cl}(A)\cap B=A\cap\mbox{cl}(B)=\emptyset .\]
A topological space $(X,\tau )$ is connected when $X$ is not the union of two nonempty separated subsets. A subset of $Y$ of $X$ is connected if and only if the topological space $Y$ with the relative topology is connected. Equivalently, $Y$ is connected if and only if $Y$ is not the union of two nonempty separated subsets. Another equivalence follows from the discussion of separation: A set $Y$ is connected if and only if the only subsets of $Y$ which are both open and closed in $Y$ are $Y$ and the empty set. From this form it follows at once that any indiscrete space is connected. A discrete space containing more than one point is not connected. The real numbers, with the usual topology, are connected, but the rationals, with the usual topology of reals relativized, are not connected. For any irrational $a$ the sets $\{x:x<a\}$ and $\{x:x>a\}$ are separated.

Proposition. Let ${\cal A}$ be a family of connected subsets of a topological space. If no two members of ${\cal A}$ are separated, then $\bigcup\{A:A\in {\cal A}\}$ is connected. $\sharp$

A component of a topological space is a maximal connected subset; that is, a connected subset which is properly contained in no other connected subset. A component of a subset $A$ is a component of $A$ with the relative topology; that is, a maximal connected subset of $A$. If a space is connected, then it is its only component. If a space is discrete, then each component consists of a single point. Of course, there are many spaces which are not discrete which have components consisting of a single point; for example, the space of rational numbers, with the (relativized) usual topology .

Proposition. Each connected subset of a topological space is contained in a component, and each component is closed. If $A$ and $B$ are distinct components of a space, then $A$ and $B$ are separated. $\sharp$

If two points, $x$ and $y$, belong to the same component of a topological space, then they always lie in the same half of a separation of the space. That is, if the space is the union of separated sets $A$ and $B$, then both $x$ and $y$ belong to $A$ or both $x$ and $y$ belong to $B$. The converse of this proposition is false. It may happen that two points always lie in the same half of a separation but nevertheless lie in different components.

\begin{equation}{\label{j}}\tag{J}\mbox{}\end{equation}

The Separation Axioms.

We shall study the topological spaces that own some separation properties.

Definition. We consider the separation axioms on a topological space $(X,\tau )$.

We say that $X$ is a $T_{1}$-space when, given any two distinct points $x$ and $y$, there exist neighborhood $N_{y}$ of $y$ satisfying $x\not\in N_{y}$,
and neighborhood $N_{x}$ of $x$ satisfying $y\not\in N_{x}$.
We say that $X$ is a $T_{2}$-space or Hausdorff space when, given any two distinct points $x$ and $y$, there exist neighborhoods $N_{x}$ and $N_{y}$ of $x$ and $y$, respectively, satisfying $N_{x}\cap N_{y}=\emptyset$.
We say that $X$ is a $T_{3}$-space when, given any closed set $F$ and any point $x\not\in F$, there exist a neighborhood $N_{x}$ of $x$ and an open set $O$ satisfying $F\subseteq O$ and $N_{x}\cap O=\emptyset$. We say that $X$ is a regular space when it is a $T_{1}$-space and $T_{3}$-space.
We say that $X$ is a $T_{4}$-space when, given any two disjoint closed sets $F_{1}$ and $F_{2}$, there exist disjoint open sets $O_{1}$ and $O_{2}$ satisfying $F_{1}\subseteq O_{1}$ and $F_{2}\subseteq O_{2}$. We say that $X$ is a normal space when it is a $T_{1}$-space and $T_{4}$-space. $\sharp$

The above all separation axioms are satisfied in metric space. We also have the following implications
\[\mbox{normal space}\Rightarrow\mbox{regular space}\Rightarrow T_{2}\Rightarrow T_{1}.\]

\begin{equation}{\label{top268}}\tag{35}\mbox{}\end{equation}
Remark \ref{top268}. From Proposition \ref{top7}, we have the following equivalences.

$X$ is a $T_{1}$-space when, given any two distinct points $x$ and $y$, there exists an open set which contains $y$ but not $x$.
$X$ is a $T_{2}$-space when, given any two distinct points $x$ and $y$, there exist disjoint open sets $O_{1}$ and $O_{2}$ satisfying $x\in O_{1}$ and $y\in O_{2}$.
$X$ is a $T_{3}$-space when, given any closed set $F$ and any point $x\not\in F$, there exist disjoint open sets $O_{1}$ and $O_{2}$ satisfying $x\in O_{1}$ and $F\subseteq O_{2}$. $\sharp$

Example. Let $X$ be a set that is totally ordered by an ordering $\prec$, and let ${\cal S}$ be the family of subset of $X$ of the form
\[\{x:x\prec a\}\mbox{ and }\{x:a\prec x\}\]
for all $a\in X$. Then, the topology $\tau$ for $X$ such that ${\cal S}$ is the subbase for $\tau$ is called the order topology induced by $\prec$. The set $X$ with the order topology is always $T_{2}$-space. The proof is given below. Given any two distinct points $a$ and $b$ in $X$, since $X$ is totally ordered, we may assume $a\prec b$.

Suppose that there exists $c\in X$ satisfying $a\prec c\prec b$. Then $\{x:x\prec c\}$ and $\{y:c\prec y\}$ are disjoint neighborhoods of $a$ and $b$, respectively.
Suppose that there is no $c\in X$ satisfying $a\prec c\prec b$. Then $\{x:x\prec b\}$ and $\{y:a\prec y\}$ are disjoint neighborhoods of $a$ and $b$,
respectively. $\sharp$

\begin{equation}{\label{top18}}\tag{36}\mbox{}\end{equation}
Proposition \ref{top18}. Let $x$ and $y$ be two distinct points of a topological space $(X,\tau )$. Then, every open set which contains either $x$ or $y$ contains both if and only if $\mbox{cl}(\{x\})=\mbox{cl}(\{y\})$.

Proof. Suppose that $\mbox{cl}(\{x\})=\mbox{cl}(\{y\})$. Since $x\in\mbox{cl}(\{x\})$, it follows $x\in\mbox{cl}(\{y\})$. Similarly, we have $y\in\mbox{cl}(\{x\})$. Let $O$ be an open set which contains $x$, but not $y$. Then $X\setminus O$ is a closed set which contains $y$, but not $x$. Therefore, we have $\mbox{cl}(\{y\})\subseteq X\setminus O$, which contradicts $x\in\mbox{cl}(\{y\})$. Therefore, there is no open set which contains $x$, but not $y$. Similarly, we can show that there is no open set which contains $y$, but not $x$.

On the other hand, suppose that every open set which contains either $x$ or $y$ contains both. Let $z\in\mbox{cl}(\{x\})$. Suppose that $O$ is an open set which contains $z$, but not $x$. Then $X\setminus O$ is a closed set which contains $x$, but not $z$. In this case, we have $\mbox{cl}(\{x\})\subseteq X\setminus O$, which says that $z\not\in\mbox{cl}(\{x\})$. This contradiction shows that every open set containing $z$ contains $x$. Since every containing $x$ contains $y$, and every open set containing $z$ contains $x$, it follows that every open set containing $z$ contains $y$, which says $z\in\mbox{cl}(\{y\})$. Therefore, we obtain the inclusion $\mbox{cl}(\{x\})\subset\mbox{cl}(\{y\})$. We can similarly obtain the inclusion $\mbox{cl}(\{y\})\subset\mbox{cl}(\{x\})$. This completes the proof. $\blacksquare$

Corollary. A topological space $(X,\tau )$ is a $T_{1}$-space if and only if, given any two distinct points $x$ and $y$ of $X$, either $x\not\in\mbox{cl}(\{y\})$ or $y\not\in\mbox{cl}(\{x\})$.

Proof. If $x\in\mbox{cl}(\{y\})$ then $\mbox{cl}(\{x\})\subset\mbox{cl}(\{y\})$, and if $y\in\mbox{cl}(\{x\})$ then $\mbox{cl}(\{y\})\subset\mbox{cl}(\{x\})$. Therefore $\mbox{cl}(\{x\})\neq\mbox{cl}(\{y\})$ if and only if either $x\not\in\mbox{cl}(\{y\})$ or $y\not\in\mbox{cl}(\{x\})$. Also $X$ is not a $T_{1}$-space if and only if every open set which contains either $x$ or $y$ contains both. The desired result follows from Proposition \ref{top18} immediately. $\blacksquare$

\begin{equation}{\label{top19}}\tag{37}\mbox{}\end{equation}
Proposition \ref{top19}. Any subspace of a $T_{1}$-space is also a $T_{1}$-space. $\sharp$

Proposition. A topological space $(X,\tau )$ is a $T_{1}$-space if and only if every set consisting of a single point is closed.

Proof. Suppose that $X$ is a $T_{1}$-space. From the first observation of Remark \ref{top268}, each $y\in X\setminus\{x\}$ is contained in an open set
$O\subseteq X\setminus\{x\}$. This says that the set $X\setminus\{x\}$ is the union of the open sets contained in it, which also means that $X\setminus\{x\}$ is open. For the converse, given two distinct points $x$ and $y$, we may take $O=X\setminus\{x\}$. Then $O$ is an open set containing $y$ but not $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top68}}\tag{38}\mbox{}\end{equation}
Proposition \ref{top68}. Any subspace of a $T_{2}$-space is also a $T_{2}$-space.

Proof. Let $Y$ be a subspace of the $T_{2}$-space $X$. Given ay two distinct $x$ and $y$of $Y$, there exist open sets $O_{x}$ and$O_{y}$ in $X$ satisfying $x\in O_{x}$, $y\in O_{y}$ and $O_{x}\cap O_{y}=\emptyset$. Since $x\in O_{x}\cap Y$, $y\in O_{y}\cap Y$ and
\[(O_{x}\cap Y)\cap (O_{y}\cap Y)=(O_{x}\cap O_{y})\cap Y=\emptyset\cap Y=\emptyset ,\]
which says that $O_{x}\cap Y$ and $O_{y}\cap Y$ are disjoint open sets in $Y$ containing $x$ and $y$, respectively. This completes the proof. $\blacksquare$

Proposition. A topological space $(X,\tau )$ is a $T_{2}$-space if and only if, given any two distinct points $x$ and $y$, there exists a neighborhood $N_{x}$ of $x$ such that $y\not\in\mbox{\em cl}(N_{x})$. $\sharp$

\begin{equation}{\label{top21}}\tag{39}\mbox{}\end{equation}
Proposition \ref{top21}. Every subspace of a $T_{3}$-space is also a $T_{3}$-space, and every subspace of a regular space is also a regular space.

Proof. Let $Y$ be a subspace of a $T_{3}$-space $(X,\tau )$. Let $F$ be closed in $Y$ and $x\in Y\setminus F$. We immediately have $F=Y\cap F^{*}$, where $F^{*}$ is closed in $X$, which also says that $x\in X\setminus F^{*}$. Since $X$ is $T_{3}$-space, there exist $O_{1},O_{2}\in\tau$ such that $x\in O_{1}$, $F^{*}\subseteq O_{2}$ and $O_{1}\cap O_{2}=\emptyset$. Then $O_{1}\cap Y$ and $O_{2}\cap Y$ are open in $Y$ such that $x\in O_{1}\cap Y$ and $F=Y\cap F^{*}\subseteq O_{2}\cap Y$. This shows that $Y$ is $T_{3}$-space. Using Proposition~\ref{top19}, we also see that every subspace of a regular space is also a regular space. This completes the proof. $\blacksquare$

\begin{equation}{\label{top22}}\tag{40}\mbox{}\end{equation}
Proposition \ref{top22}. A topological space $(X,\tau )$ is a $T_{3}$-space if and only if, given any $x\in O_{x}\in\tau$, there exists $O_{x}^{*}\in\tau$ containing $x$ satisfying $\mbox{cl}(O_{x}^{*})\subseteq O_{x}$.

Proof. We first have that $X\setminus O_{x}$ is closed in $X$ and does not contain $x$. The $T_{3}$-space says that there exist $O_{x}^{*},O_{x}^{\circ}\in\tau$ satisfying $X\setminus O_{x}\subseteq O_{x}^{\circ}$, $x\in O_{x}^{*}$ and $O_{x}^{\circ}\cap O_{x}^{*}=\emptyset$. Since $X\setminus O_{x}\subseteq O_{x}^{\circ}$, it also says $X\setminus O_{x}^{\circ}\subseteq O_{x}$. Since $O_{x}^{\circ}\cap O_{x}^{*}=\emptyset$, we have
\[O_{x}^{*}\subseteq X\setminus O_{x}^{\circ}\subseteq O_{x}.\]
Since $X\setminus O_{x}^{\circ}$ is closed, we obtain
\[\mbox{cl}(O_{x}^{*})\subseteq X\setminus O_{x}^{\circ}\subseteq O_{x}.\]

For the converse, given any $x\in X$ and any closed set $F$ with $x\not\in F$, it says that $X\setminus F$ is open containing $x$. The hypothesis says that there exists an open set $O_{x}^{*}\in\tau$ containing $x$ such that $\mbox{cl}(O_{x}^{*})\subset X\setminus F$, which says that $F\subseteq X\setminus\mbox{cl}(O_{x}^{*})\in\tau$. Since $O_{x}^{*}\subset\mbox{cl}(O_{x}^{*})$, it follows
\[[X\setminus\mbox{cl}(O_{x}^{*})]\cap O_{x}^{*}=\emptyset ,\]
which says that $X\setminus\mbox{cl}(O_{x}^{*})$ and $O_{x}^{*}$ are open in $X$ separating $F$ and $x$, respectively. This completes the proof.$\blacksquare$

Proposition. A topological space $(X,\tau )$ is a $T_{4}$-space if and only if, given any closed set $F$ and any open set $O$ with $F\subseteq O$, there exists an open set $O^{*}$ satisfying
\[F\subseteq O^{*}\subset\mbox{cl}(O^{*})\subseteq O.\]

Proof. Suppose that $X$ is $T_{4}$-space. We see that $X\setminus O$ is closed and $(X\setminus O)\cap F=\emptyset$. The $T_{4}$-space says that the exist open sets $O^{*}$ and $O^{\circ}$ such that $X\setminus O\subseteq O^{\circ}$, $F\subseteq O^{*}$ and $O^{\circ}\cap O^{*}=\emptyset$. Since $X\setminus O\subseteq O^{\circ}$, it also says that $X\setminus O^{\circ}\subseteq O$. Since $O^{\circ}\cap O^{*}=\emptyset$, we have
\[O^{*}\subseteq X\setminus O^{\circ}\subseteq O.\]
Since $X\setminus O^{\circ}$ is closed, we obtain
\[F\subseteq O^{*}\subset\mbox{cl}(O^{*})\subseteq X\setminus O^{\circ}\subseteq O.\]

For the converse, given any disjoint closed sets $F_{1}$ and $F_{2}$, it says that $X\setminus F_{1}$ is open containing $F_{2}$. The hypothesis says that there exists an open set $O^{*}\in\tau$ satisfying
\[F_{2}\subseteq O^{*}\subset\mbox{cl}(O^{*})\subseteq X\setminus F_{1},\]
which says $F_{1}\subseteq X\setminus\mbox{cl}(O^{*})\in\tau$. Since $O^{*}\subset\mbox{cl}(O^{*})$, it follows
\[[X\setminus\mbox{cl}(O^{*})]\cap O^{*}=\emptyset ,\]
which says that $X\setminus\mbox{cl}(O^{*})$ and $O^{*}$ are open in $X$ separating $F_{1}$ and $F_{2}$, respectively. This completes the proof. $\blacksquare$

We need to remark that every subspace of a $T_{4}$-space is not necessarily a $T_{4}$-space. However, we have the following weaker statement.

\begin{equation}{\label{top25}}\tag{41}\mbox{}\end{equation}
Proposition \ref{top25}. If $Y$ is a subspace of a $T_{4}$-space $(X,\tau )$ and is also closed in $X$, then $Y$ is a $T_{4}$-space, and if $Y$ is a subspace of a normal space $(X,\tau )$ and is also closed in $X$, then $Y$ is a normal space.

Proof. Suppose that $Y$ is closed in $X$ and is a subspace. If $F$ is closed in $Y$ then $F=Y\cap F^{*}$, where $F^{*}$ is closed in $X$, which also says that $F$ is closed in $X$. In other words, if $F_{1}$ and $F_{2}$ are closed in $Y$, then they are closed in $X$. The $T_{4}$-space says that there exist open sets $O_{1},\O_{2}\in\tau$ satisfying $F_{1}\subseteq O_{1}$, $F_{2}\subseteq O_{2}$ and $O_{1}\cap O_{2}=\emptyset$, which imply $F_{1}\subseteq Y\cap O_{1}$, $F_{2}\subseteq Y\cap O_{2}$ and
\[(Y\cap O_{1})\cap (Y\cap O_{2})=(O_{1}\cap O_{2})\cap Y=\emptyset .\]
Therefore, we conclude that $Y\cap O_{1}$ and $Y\cap O_{2}$ are disjoint open sets in $Y$ separating $F_{1}$ and $F_{2}$, respectively. This completes the proof. $\blacksquare$

\begin{equation}{\label{royl85}}\tag{42}\mbox{}\end{equation}
Lemma \ref{royl85} (Urysohn’s Lemma). Let $A$ and $B$ be disjoint closed subsets of a normal space $X$. Then, there is a continuous real-valued function $f$ defined on $X$ such that $0\leq f \leq 1$ on $X$ while $f\equiv 0$ on $A$ and $f\equiv 1$ on $B$. $\sharp$

\begin{equation}{\label{top26}}\tag{43}\mbox{}\end{equation}
Lemma \ref{top26}. (Urysohn’s Lemma).
A topological space $(X,\tau )$ is a $T_{4}$-space if and only if, given any two disjoint closed sets $A$ and $B$ of $X$, there exists a continuous function $f$ from $X$ into $[0,1]$ that is endowed with the usual topology induced by the absolute value metric satifying $f(a)=0$ for all $a\in A$ and $f(b)=1$ for all $b\in B$. $\sharp$

Corollary. Suppose that $(X,\tau )$ is a normal space which contains more than one point. Then, there exists a nonconstant continuous function $f$ from $X$ into $[0,1]$.

Proof. Since $X$ is $T_{1}$-space, each singleton set is closed. Given any two distinct points $a$ and $b$ of $X$. Then $\{a\}$ and $\{b\}$ are closed sets in $X$. Urysohn’s Lemma \ref{top26} says that there exists a continuous function $f$ from $X$ into $[0,1]$ satisfying $f(a)=0$ and $f(b)=1$. This completes the proof. $\blacksquare$

Theorem. (Tietze’s Extension Theorem).
Let $(X,\tau )$ be a normal topological space, and let $Y$ be a closed subset of $X$. Suppose that $f$ is a continuous function from $Y$ into $\mathbb{R}$ that is endowed with the usual topology induced by the absolute value metric. Then, there is a continuous extension $F$ of $f$ from $X$ into $\mathbb{R}$; that is, $F|_{Y}=f$ or $F(x)=f(x)$ for $x\in Y$. $\sharp$

We see that Urysohn’s Lemma is a special case of the Tietze’s Extension Theorem. Let $A$ and $B$ be disjoint closed sets. Then, we can define $Y=A\cup B$ and $f:Y\rightarrow\mathbb{R}$ by $f(a)=0$ for all $a\in A$ and $f(b)=1$ for all $b\in B$. The Tietze’s Extension Theorem says that $f$ can be extended to a continuous function $F$, which is the statement of Urysohn’s Lemma. However, the proof of the Tietze’s Extension Theorem needs to invoke the Urysohn’s Lemma.

Corollary. Let $(X,\tau )$ be a normal topological space, and let $Y$ be a closed subset of $X$. Suppose that $\hat{\bf f}=(f_{1},\cdots ,f_{n})$ is a continuous function on $Y$ into $\mathbb{R}^{n}$. Then, there is a continuous extension ${\bf F}$ of $\hat{\bf f}$ from $X$ into $\mathbb{R}^{n}$.

Proof. Let $p_{i}$ be the projection from $\mathbb{R}^{n}$ into the $i$th coordinate. Then $f_{i}\equiv p_{i}\circ f$ is a continuous function from $Y$ into $\mathbb{R}$. The Tietze’s Extension Theorem says that each $f_{i}$ has a continuous extension $F_{i}$ for all $i=1,\cdots ,n$. Then, we define ${\bf F}$ by ${\bf F}(x)=(F_{1}(x),\cdots ,F_{n}(x))$ for $x\in X$. Then ${\bf F}$ is also continuous. This completes the proof. $\blacksquare$

Theorem. (Urysohn Metrization Theorem). Every normal topological space satisfying the second axiom of countability is metrizable. $\sharp$

Two subsets $A$ and $B$ are separated in a topological space $X$ if and only if $\mbox{cl}(A)\cap B$ and $A\cap \mbox{cl}(B)$ are both empty sets. This definition of separation invloves the closure operation in $X$. However, the apparent dependence on the space $X$ is illusory, for $A$ and $B$ are separated in $X$ if and only if neither $A$ nor $B$ contains a point or an accumulation point of other. This condition may be restated in terms of of the relative topology for $A\cup B$, in view of part (ii) of the foregoing Proposition~\ref{top13}, as: both $A$ and $B$ are closed in $A\cup B$ (or equivalently $A$ (or $B$) is both open and closed in $A\cup B$) and $A$ and $B$ are disjoint. As an example, notice that the open intervals $(0,1)$ and $(1,2)$ are separated subsets of the real numbers with the usual topology and that there is a point, $1$, belonging to the closure of both. However, $(0,1)$ is not separated from the closed interval $[1,2]$ because $1$, which is a member of $[1,2]$, is an accumulation point of $(0,1)$.

Proposition. If $Y$ and $Z$ are subsets of a topological space $X$ and both $Y$ and $Z$ are closed or both are open, then $Y\setminus Z$ is separated from $Z\setminus Y$. $\sharp$

Proposition. Let $X$ be a topological space which is the union of subsets $Y$ and $Z$ such that $Y\setminus Z$ and $Z\setminus Y$ are separated. Then, the closure of a subset $A$ of $X$ is the union of the closure in $Y$ of $A\cap Y$ and the closure in $Z$ of $A\cap Z$. $\sharp$

Corollary. Let $X$ be a topological space which is the union of subsets $Y$ and $Z$ such that $Y\setminus Z$ and $Z\setminus Y$ are separated. Then a subset $A$ of $X$ is closed (open) if $A\cap Y$ is closed (open) in $Y$ and $A\cap Z$ is closed (open) in $Z$. $\sharp$

\begin{equation}{\label{k}}\tag{K}\mbox{}\end{equation}

Convergence.

Definition. A sequence is a function defined on $\mathbb{N}$ into a universal set $X$.

We say that a sequence $\{x_{n}\}_{n=1}^{\infty}$ is in a set $A$ when $x_{n}\in A$ for each $n$.
We say that a sequence $\{x_{n}\}_{n=1}^{\infty}$ is eventually in a set $A$ when there exists an integer $n_{0}$ satisfying $x_{n}\in A$ for $n\geq n_{0}$.
We say that a sequence $\{x_{n}\}_{n=1}^{\infty}$ is frequently in a set $A$ when, for each nonnegative integer $n_{0}$, there exists an integer $n$ satisfying $x_{n}\in A$ for $n\geq n_{0}$. $\sharp$

Definition. Suppose that $X$ is a topological space.

We say that a sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ converges to the point $x\in X$ when, given any neighborhood $N$ of $x$, there exists an integer $n_{0}$ satisfying $x_{n}\in N$ for all $n\geq n_{0}$.
We say that a sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ has a cluster point $x\in X$ when, given any neighborhood $N$ of $x$ and any integer $n$, there exists an integer $n_{0}\geq n$ satisfying $x_{n_{0}}\in N$. $\sharp$

We have the following observations.

A sequence converges to a point $x$ if and only if it is eventually in each neighborhood of $x$.
A sequence has a cluster point $x$ if and only if it is frequently in each neighborhood of $x$.
A point $x$ belongs to the closure of $A$ if and only if there is a sequence in $A$ which converges to $x$.
Suppose that the sequence $\{x_{n}\}_{n=1}^{\infty}$ is frequently in $A$. It is equivalent to say that the sequence $\{x_{n}\}_{n=1}^{\infty}$ is not eventually in the complement of $A$. Intuitively, a sequence is frequently in $A$ if it keeps returning to $A$.

We shall now construct subsequences of a sequence. A sequence $\{x_{n}\}_{n=1}^{\infty}$ may converge to no point. We wish to select an integer $n_{j}$, for each nonnegative integer $j$, such that the sequence $\{x_{n_{j}}\}_{j=1}^{\infty}$ is convergent. The formal definition is given below.

Definition. Let $\{x_{i}\}_{i=1}^{\infty}$ be a sequence. We say that $\{y_{j}\}_{j=1}^{\infty}$ is a subsequence of $\{x_{i}\}_{i=1}^{\infty}$ when there is a sequence $\{n_{j}\}_{j=1}^{\infty}$ of nonnegative integers satisfying $y_{j}=x_{n_{j}}$ for each $j$, and for each integer $n_{0}$ there exists an integer $n_{1}$ such that $j\geq n_{1}$ implies $n_{j}\geq n_{0}$. $\sharp$

We have the following observations.

If a sequence is eventually in a set, then every subsequence is also eventually in the same set.
If a sequence converges to a point, then every subsequence also converges to the same point.
If the sequence $\{x_{n}\}_{n=1}^{\infty}$ has a subsequence which converges to $x$, then $x$ is a cluster point of $\{x_{n}\}_{n=1}^{\infty}$.
The converse statement is not always true.

Definition. A directed set is a pair $(\Gamma ,\prec )$ such that the following conditions are satisfied:

if $\gamma\in\Gamma$, then $\gamma\prec\gamma$;
if $\gamma\prec\beta$ and $\beta\prec\gamma$, then $\gamma\prec\gamma$;
if $\alpha ,\beta\in\Gamma$, there exists $\gamma\in\Gamma$ satisfying $\alpha\prec\gamma$ and $\beta\prec\gamma$.

We say that $\gamma$ follows $\beta$ and that $\beta$ precedes $\gamma$ when $\gamma\succ\beta$. $\sharp$

Example. We have many examples regarding the directed sets.

The set $\mathbb{N}$ of positive integers with $\prec$ replaced by $\leq$ is a directed set.
The set of all open sets containing a point $x$ with $O_{1}\prec O_{2}$ defined to mean $O_{2}\subseteq O_{1}$ is a directed set, since the intersection of two open sets containing $x$ is also an open set containing $x$. $\sharp$

\begin{equation}{\label{top130}}\tag{44}\mbox{}\end{equation}
Definition \ref{top130}. Let $(X,\tau )$ be a topological space.

A {\bf net} $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is a mapping of a directed set $\Gamma$ into $X$. If the directed set is the set of positive integers,
then the net is also called a sequence.
A net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is eventually in a set $A$ when there exists $\gamma_{0}\in\Gamma$ such that $\gamma\in\Gamma$ and $\gamma\succ\gamma_{0}$ imply $x_{\gamma}\in A$.
A net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in a set $A$ when, for each $\gamma\in\Gamma$, there exists $\gamma_{0}\in\Gamma$ satisfying $\gamma_{0}\succ\gamma$ and $x_{\gamma_{0}}\in A$.
A point $x\in X$ is said to be the limit of a net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ when, given any neighborhood $N$ of $x$, there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in N_{x}$ for all $\gamma\succ\gamma_{0}$. In this case, we say that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$.
A point $x$ is called a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ when, given any neighborhood $N$ of $x$ and any $\gamma\in\Gamma$, there exists $\beta\succ\gamma$ satisfying $x_{\beta}\in N$. $\sharp$

\begin{equation}{\label{top38}}\tag{45}\mbox{}\end{equation}
Remark \ref{top38}. We have the following observations.

The net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$ if and only if the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is eventually in each neighborhood $N$ of $x$.
The net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a cluster point $x$ if and only if the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in each neighborhood $N$ of $x$. $\sharp$

Proposition. Let $(X,\tau )$ be a topological space. Then, we have the following properties.

(i) The net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in $X$ converges to $x$ if and only if, given any open set $O$ containing $x$, there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in O$ for all $\gamma\succ\gamma_{0}$. In particular, the sequence $\{x_{n}\}_{n=1}^{\infty}$ converges to $x$ if and only if, given any open set $O$ containing $x$, there exists an integer $n_{0}$ satisfying $x_{n}\in O$ for all $n\geq n_{0}$.

(ii) A point $x$ is a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in $X$ if and only if, given any open set $O$ containing $x$ and any $\gamma\in\Gamma$, there exists $\beta\succ\gamma$ satisfying $x_{\beta}\in O$. In particular, a point $x$ is a cluster point of the sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ if and only if, given any open set $O$ containing $x$ and any integer $n$, there exists an integer $n_{0}\geq n$ satisfying $x_{n_{0}}\in O$.

(iii) Let ${\cal B}$ be a base for $\tau$. The net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in $X$ converges to $x$ if and only if, given any $B\in {\cal B}$ containing $x$, there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in B$ for all $\gamma\succ\gamma_{0}$. In particular, the sequence $\{x_{n}\}_{n=1}^{\infty}$ converges to $x$ if and only if, given any $B\in {\cal B}$ containing $x$, there exists an integer $n_{0}$ satisfying $x_{n}\in B$
for all $n\geq n_{0}$.

(iv) Let ${\cal B}$ be a base for $\tau$. A point $x$ is a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in $X$ if and only if, given any $B\in {\cal B}$ containing $x$ and any $\gamma\in\Gamma$, there exists $\beta\succ\gamma$ satisfying $x_{\beta}\in B$. In particular, a point $x$ is a cluster point of the sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ if and only if, given any $B\in {\cal B}$ containing $x$ and any integer $n$, there exists an integer $n_{0}\geq n$ satisfying $x_{n_{0}}\in B$.

Proof. To prove part (i), suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. Since any open set $O$ containing $x$ is a neighborhood of $x$, the convergence says that there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in O$ for all $\gamma\succ\gamma_{0}$.
For the converse, given any neighborhood $N$ of $x$, there exists an open set $O$ containing $x$ satisfying $x\in O\subseteq N$. The assumption says that there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in O\subseteq N$ for all $\gamma\succ\gamma_{0}$, which shows the convergence of the net. Part (ii) can be similarly obtained.

To prove part (iii), suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. Since any $B\in {\cal B}$ containing $x$ is also an open set containing $x$, the convergence says that there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in B$ for all $\gamma\succ\gamma_{0}$. For the converse, given any open set $O$ containing $x$, there exists $B\in {\cal B}$ satisfying $x\in B\subseteq O$. The assumption says that there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma}\in B\subseteq O$ for all $\gamma\succ\gamma_{0}$, which shows the convergence of the net. Part (iv) can be similarly obtained. This completes the proof. $\blacksquare$

\begin{equation}{\label{top264}}\tag{46}\mbox{}\end{equation}
Proposition \ref{top264}. Let $A$ be a subset set of a topological space $X$. Then, we have the following properties.

(i) A point $x$ is an accumulation point of $A$ if and only if there is a net in $A\setminus\{x\}$ which converges to $x$.

(ii) A point $x$ belongs to the closure of $A$ if and only if there is a net in $A$ which converges to $x$.

Proof. To prove part (i), if $x$ is an accumulation point of $A$, then for each neighborhood $U$ of $x$, there exists a point $x_{U}$ of $A$ which belongs to $U\setminus\{x\}$. The family ${\cal N}_{x}$ of all neighborhoods of $x$ is directed by $\subseteq$. If $U$ and $V$ are neighborhoods of $x$ satisfying $V\subseteq U$, i.e., $U\prec V$. Now, we consider the net $\{x_{U}\}_{U\in {\cal U}}$. Given any neighborhood $N$ of $x$, we take $U_{0}=N$. Then, for any $U\succ U_{0}$, i.e., $U\subseteq U_{0}=N$, we have $x_{U}\in U\setminus\{x\}\subseteq N$, which shows that the net $\{x_{U}\}_{U\in {\cal U}}$ converges to $x$. For the converse, if a net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in $A\setminus\{x\}$ converges to $x$, then, for each neighborhood $N$ of $x$,
\[\{x_{\gamma}\}_{\gamma\in\Gamma}\cap (N\setminus\{x\})\neq\emptyset,\]
which shows that $x$ is an accumulation point of $A$.

To prove part (ii), since the closure of a set $A$ consists of $A$ together with all the accumulation points of $A$. For any $x\in A$, we can take a net whose elements are all $x$. For each accumulation point $x$ of $A$, part (i) says that there exists a net in $A$ converging to $x$. For the converse, if there exists a net in $A$ converging to $s$, then every neighborhood of $x$ intersects $A$, which says that $x$ belongs to the closure of $A$. This completes the proof. $\blacksquare$

Proposition. Let $X$ be a topological space satisfying the first axiom of countability. Then, we have the following properties.

(i) A point $x$ is an accumulation point of a subset $A$ of $X$ if and only if there is a sequence in $A\setminus\{x\}$ which converges to $x$.

(ii) A point $x$ belongs to the closure of a subset $A$ of $X$ if and only if there is a sequence in $A$ which converges to $x$.

Proof. To prove part (i), let $x$ be an accumulation point of $A$. Suppose that $\{N_{n}\}_{n=1}^{\infty}$ is a countable neighborhood base at $x$. We define $V_{n}=\bigcap_{i=1}^{n}N_{i}$. Then $\{V_{n}\}_{n=1}^{\infty}$ is also a neighborhood base at $x$ satisfying $V_{n+1}\subseteq V_{n}$ for each $n$. Since $V_{n}\cap (A\setminus\{x\})\neq\emptyset$, we can take $x_{n}\in V_{n}$ with $x_{n}\neq x$. Given any neighborhood $N$ of $x$, there exists $n_{0}$ satisfying $V_{n_{0}}\subseteq N$. Therefore, for $n\geq n_{0}$, we have
\[x_{n}\in V_{n}\subseteq V_{n_{0}}\subseteq N,\]
which shows that the sequence $\{x_{n}\}_{n=1}^{\infty}$ converges to $x$. For the converse, if a sequence $\{x_{n}\}_{n=1}^{\infty}$ in $A\setminus\{x\}$ converges to $x$, then, for each neighborhood $N$ of $x$,
\[\{x_{n}\}_{n=1}^{\infty}\cap (N\setminus\{x\})\neq\emptyset,\]
which shows that $x$ is an accumulation point of $A$. Part (ii) can be similarly obtained according to the argument of part (ii) of Proposition \ref{top264}. This completes the proof. $\blacksquare$

We note that a net may converge to several different points. For example, if $(X,\tau )$ is a trivial topology, then any sequence $\{x_{n}\}_{n=1}^{\infty}$ converges to every point of $X$, since if $x\in X$ then the only neighborhood of $x$ is $X$.

Theorem. A topological space is a Hausdorff space if and only if each net in the space converges to at most one point.

Proof. Let $(X,\tau )$ be a Hausdorff space. Given any two distinct points $x$ and $y$ in $X$, there exists disjoint neighborhoods $N_{x}$ and $N_{y}$ of $x$ and $y$, respectively. Suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$ and $y$. Then, there exist $\alpha_{0},\beta_{0}\in\Gamma$ satisfying $x_{\gamma}\in N_{x}$ for $\alpha\succ\alpha_{0}$ and $x_{\beta}\in N_{y}$ for $\beta\succ\beta_{0}$. The definition of directed set says that there exists $\gamma_{0}\in\Gamma$ satisfying $\gamma_{0}\succ\alpha_{0}$ and $\gamma_{0}\succ\beta_{0}$. Therefore, we must have $x_{\gamma_{0}}\in N_{x}\cap N_{y}$, which contradicts $N_{x}\cap N_{y}=\emptyset$. This says that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to at most one point.

For the converse, assume that $X$ is not a Hausdorff space. Therefore, there exist two distinct points $x$ and $y$ such that every neighborhood of $x$ intersects every neighborhood of $y$. Let ${\cal N}_{x}$ be the family of all neighborhoods of $x$, and let ${\cal N}_{y}$ be the family of all neighborhoods of $y$. Then $N_{x}\cap N_{y}\neq\emptyset$ for any $N_{x}\in {\cal N}_{x}$ and $N_{y}\in {\cal N}_{y}$. We also see that both ${\cal N}_{x}$ and ${\cal N}_{y}$ are directed by $\subseteq$. Considering the Cartesian product ${\cal N}_{x}\times {\cal N}_{y}$, we define the ordering
\[\left (N_{x}^{(1)},N_{y}^{(1)}\right )\succ\left (N_{x}^{(2)},N_{y}^{(2)}\right )\mbox{ if and only if }N_{x}^{(1)}\subseteq N_{x}^{(2)}\mbox{ and }
N_{y}^{(1)}\subseteq N_{y}^{(2)}.\]
It is clear to see that the Cartesian product is directed by $\succ$. For each $(N_{x},N_{y})\in {\cal N}_{x}\times {\cal N}_{y}$, we have $N_{x}\cap N_{y}\neq\emptyset$. In this case, we can select a point $s_{(N_{x},N_{y})}$ from $N_{x}\cap N_{y}$. Therefore,
\[\left (N_{x}^{(1)},N_{y}^{(1)}\right )\succ\left (N_{x}^{(2)},N_{y}^{(2)}\right )\mbox{ implies }
s_{(N_{x}^{(1)},N_{y}^{(1)})}\in\left (N_{x}^{(1)}\cap N_{y}^{(1)}\right )\subseteq
\left (N_{x}^{(2)}\cap N_{y}^{(2)}\right ).\]
Given any neighborhoods $N_{x}^{*}$ of $x$ and $N_{y}^{*}$ of $y$, if $(N_{x},N_{y})\succ (N_{x}^{*},N_{y}^{*})$, then we have
\[s_{(N_{x},N_{y})}\in\left (N_{x}\cap N_{y}\right )\subseteq\left (N_{x}^{*}\cap N_{y}^{*}
\right ),\mbox{ i.e., }s_{(N_{x},N_{y})}\in N^{*}_{x}\mbox{ and }s_{(N_{x},N_{y})}\in N^{*}_{y}.\]
Therefore, the net
\[\{s_{(N_{x},N_{y})}\}_{(N_{x},N_{y})\in {\cal N}_{x}\times {\cal N}_{y}}\]
converges to both $x$ and $y$. This contradiction completes the proof. $\blacksquare$

\begin{equation}{\label{top37}}\tag{47}\mbox{}\end{equation}
Definition \ref{top37}. Let $\Gamma$ and $\Lambda$ be two directed sets. A net $\{y_{\beta}\}_{\beta\in\Gamma}$ is a subnet of a net $\{x_{\gamma}\}_{\gamma\in\Lambda}$ when there exists a function $\eta:\Gamma\rightarrow\Lambda$ written as $\eta_{\beta}=\eta(\beta )\in\Lambda$ such that the following conditions are satisfied:

$y_{\beta}=x_{\eta_{\beta}}$ for each $\beta\in\Gamma$;
for each $\gamma\in\Lambda$, there exists $\beta\in\Gamma$ such that $\gamma\in\Lambda$ and $\gamma\succ\beta$ imply $\eta_{\gamma}\succ\gamma$.

The second condition says that if $\gamma$ becomes large then $\eta_{\gamma}$ also becomes large. $\sharp$

\begin{equation}{\label{top30}}\tag{48}\mbox{}\end{equation}
Proposition \ref{top30}. Suppose that $N$ is a function from the direct set $\Gamma$ into the directed set $\Gamma$ such that the following conditions are satisfied.

$\beta_{1}\succ\beta_{2}$ implies $N_{\beta_{1}}\succ N_{\beta_{2}}$.
The range of $N$ is cofinal in $\Gamma$; that is, for each $\gamma\in\Gamma$. there is $\beta\in\Gamma$ satisfying $N_{\beta}\succ\gamma$; equivalently, if $\gamma_{1},\gamma_{2}\in\Gamma$, then there exists $\beta\in\Gamma$ satisfying $\gamma_{1}\prec N_{\beta}$ and $\gamma_{2}\prec N_{\beta}$.

Then $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ for each net $\{x_{\gamma}\}_{\gamma\in\Gamma}$.

Proof. We first present the equivalence for the second condition. Suppose that $\gamma_{1},\gamma_{2}\in\Gamma$. Then, there exists $\gamma\in\Gamma$ satisfying $\gamma\succ\gamma_{1}$ and $\gamma\succ\gamma_{2}$. The second condition says that there exists $\beta\in\Gamma$ satisfying $N_{\beta}\succ\gamma$, which also implies $N_{\beta}\succ\gamma_{1}$ and $N_{\beta}\succ\gamma_{2}$. The converse is obvious. Now, given any net $\{x_{\gamma}\}_{\gamma\in\Gamma}$, for each $\gamma\in\Gamma$, the second condition says that there is $\beta\in\Gamma$ satisfying $N_{\beta}\succ\gamma$. If $\gamma\in\Gamma$ and $\gamma\succ\beta$, then the first condition says $N_{\gamma}\succ N_{\beta}$, which implies $N_{\gamma}\succ\gamma$. This shows that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet,
and the proof is complete. $\blacksquare$

\begin{equation}{\label{top36}}\tag{49}\mbox{}\end{equation}
Proposition \ref{top36}. We have the following properties.

(i) If the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is eventually in a set $A$, then the subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is also eventually in $A$.

(ii) If the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$, then every subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ also converges to $x$

Proof. To prove part (i), since the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is eventually in $A$, there exists $\gamma_{0}\in\Gamma$ such that $\gamma\in\Gamma$ and $\gamma\succ\gamma_{0}$ imply $x_{\gamma}\in A$. From the second condition of Definition \ref{top37}, there exists $\beta_{0}\in\Gamma$ such that if $\beta\in\Gamma$ and $\beta\succ\beta_{0}$, then $N_{\beta}\succ\gamma_{0}$, which also says $x_{N_{\beta}}\in A$. Part (ii) follows from part (i) and the first observation of Remark \ref{top38}. This completes the proof. $\blacksquare$

\begin{equation}{\label{top35}}\tag{50}\mbox{}\end{equation}
Proposition \ref{top35}. If every subnet of a net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a subsubnet which converges to $x$, then the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. Equivalently, if the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ does not converge to $x$, then there exists a subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ such that no subnet of $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ converges to $x$.

Proof. Suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ does not converge to $x$. Then, there exists a neighborhood $N_{x}$ of $x$ such that there does not exist any $\gamma_{0}\in\Gamma$ such that $\gamma\succ\gamma_{0}$ implies $x_{\gamma}\in N_{x}$. We define the index set
\[\Gamma =\left\{\gamma\in\Gamma :x_{\gamma}\not\in N_{x}\right\}.\]
Let $N$ be the identity function from $\Gamma\rightarrow\Gamma$. Then $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$. Since each $x_{N_{\beta}}\not\in N_{x}$, any subnet of $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ cannot converge to $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top53}}\tag{51}\mbox{}\end{equation}
Proposition \ref{top53}. Let $\{x_{\gamma}\}_{\gamma\in\Gamma}$ be a net in the topological space. Then $x$ is a cluster point of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ if and only if $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a subnet that converges to $x$.

Proof. Suppose that $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a subnet which converges to $x$. Then, there exists a directed set $\Gamma$ and a function $N:\Gamma\rightarrow\Gamma$ such that the net $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ converges to $x$. Given any neighborhood $N_{x}$ of $x$, there exists $\beta_{0}\in\Gamma$ such that $\beta\succ\beta_{0}$ implies $x_{N_{\beta}}\in N_{x}$. Suppose that $\gamma_{1},\gamma_{2}\in\Gamma$. Since $\Gamma$ is a directed set, there exists $\gamma_{0}\in\Gamma$ satisfying $\gamma_{1}\prec\gamma_{0}$ and $\gamma_{2}\prec\gamma_{0}$. By the second condition of Definition \ref{top37}, there exists $\beta_{1}\in\Gamma$ such that $\gamma\succ\beta_{1}$ implies $N_{\gamma}\succ\gamma_{0}$. Since $\Gamma$ is a directed set, there exists $\beta_{2}\in\Gamma$ satisfying $\beta_{0}\prec\beta_{2}$ and $\beta_{1}\prec\beta_{2}$, which also says $x_{N_{\beta_{2}}}\in N_{x}$ and $N_{\beta_{2}}\succ\gamma_{0}$. The transitivity says $\gamma_{1}\prec N_{\beta_{2}}$ and $\gamma_{2}\prec N_{\beta_{2}}$. Therefore, we can say that, given any neighborhood $N_{x}$ of $x$ and any $\gamma\in\Gamma$, there exists $N_{\beta_{2}}\in\Gamma$ satisfying $\gamma\prec N_{\beta_{2}}$ and $x_{N_{\beta_{2}}}\in N_{x}$. This says that $x$ is a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$.

Now suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a cluster point $x$. Let
\[\Gamma =\left\{(\gamma ,N_{x}):\mbox{$N_{x}$ is a neighborhood of $x$ and $x_{\gamma}\in N_{x}$}\right\}.\]
Then $\Gamma$ is ordered by defining
\[(\gamma_{1},N_{x}^{(1)})\prec (\gamma_{2},N_{x}^{(2)})\mbox{ if and only if }
\gamma_{1}\prec\gamma_{2}\mbox{ and }N_{x}^{(2)}\subseteq N_{x}^{(1)}.\]
It is easy to show that $\Gamma$ is partially ordered. Moreover, we want to show that $\Gamma$ is a directed set. Given any $(\gamma_{1},N_{x}^{(1)})$ and $(\gamma_{2},N_{x}^{(2)})$, we see that $N_{x}^{(1)}\cap N_{x}^{(2)}$ is a neighborhood of $x$ with
\[N_{x}^{(1)}\cap N_{x}^{(2)}\subseteq N_{x}^{(1)}\mbox{ and }N_{x}^{(1)}\cap N_{x}^{(2)}\subseteq N_{x}^{(2)}.\]
Since $\Gamma$ is a directed set, there exists $\gamma_{3}\in\Gamma$ such that $\gamma_{1}\prec\gamma_{3}$ and $\gamma_{2}\prec\gamma_{3}$. By the definition of cluster point, there exists $\gamma_{0}\in\Gamma$ satisfying $x_{\gamma_{0}}\in N_{x}^{(1)}\cap N_{x}^{(2)}$ for $\gamma_{3}\prec\gamma_{0}$. The transitivity also says $\gamma_{1}\prec\gamma_{0}$ and $\gamma_{2}\prec\gamma_{0}$. Therefore $(\gamma_{0},N_{x}^{(1)}\cap N_{x}^{(2)})\in\Gamma$ such that we have
\[(\gamma_{1},N_{x}^{(1)})\prec (\gamma_{0},N_{x}^{(1)}\cap N_{x}^{(2)})\mbox{ and }
(\gamma_{2},N_{x}^{(2)})\prec (\gamma_{0},N_{x}^{(1)}\cap N_{x}^{(2)}),\]
which shows that $\Gamma$ is a directed set. Define the function $N$ from $\Gamma$ into $\Gamma$ by $N(\gamma ,N_{x})=\gamma$. We want to claim that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$. For each $\gamma\in\Gamma$ and any
neighborhood $N_{x}$ of $x$, since $x$ is a cluster point, there exists $\gamma_{0}\in\Gamma$ satisfying $\gamma\prec\gamma_{0}$ and $x_{\gamma_{0}}\in N_{x}$, which says $(\gamma_{0},N_{x})\in\Gamma$. If $(\gamma^{*},N_{x}^{*})\succ (\gamma_{0},N_{x})$, i.e., $\gamma^{*}\succ\gamma_{0}$ and $N_{x}^{*}\subseteq N_{x}$, then
\[N(\gamma^{*},N_{x}^{*})=\gamma^{*}\succ\gamma_{0}\succ\gamma .\]
This says $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$. Finally, we want to show that the subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ converges to $x$. Given any neighborhood $N_{x}$ of $x$ and any $\gamma\in\Gamma$, since $x$ is a cluster point, there exists $\gamma_{0}\in\Gamma$ satisfying $\gamma\prec\gamma_{0}$ and $x_{\gamma_{0}}\in N_{x}$, which says $(\gamma_{0},N_{x})\in\Gamma$. If $(\gamma^{*},N_{x}^{*})\succ (\gamma_{0},N_{x})$, i.e., $N_{x}^{*}\subseteq N_{x}$ and
$x_{\gamma^{*}}\in N_{x}^{*}$, then
\[x_{N(\gamma^{*},N_{x}^{*})}=x_{\gamma^{*}}\in N_{x}^{*}\subseteq N_{x}.\]
In other words, given any neighborhood $N_{x}$ of $x$, there exists $(\gamma_{0},N_{x})\in\Gamma$ such that $(\gamma^{*},N_{x}^{*})\succ (\gamma_{0},N_{x})$ implies $x_{N(\gamma^{*},N_{x}^{*})}\in N_{x}$, which shows that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$
converges to $x$. This completes the proof. $\blacksquare$

Corollary. Suppose that a subnet of the $\{x_{\gamma}\}_{\gamma\in\Gamma}$ has a cluster point $x$. Then $x$ is also a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$. $\sharp$

Proposition. Let $X$ be a topological space satisfying the first axiom of countability. If $x$ is a cluster point of a sequence $\{x_{n}\}_{n=1}^{\infty}$ then there exists a subsequence of $\{x_{n}\}_{n=1}^{\infty}$ converging to $x$.

Proof. Suppose that $\{N_{n}\}_{n=1}^{\infty}$ is a countable neighborhood base at $x$. We define $V_{n}=\bigcap_{i=1}^{n}N_{i}$. Then $\{V_{n}\}_{n=1}^{\infty}$ is also a neighborhood base at $x$ satisfying $V_{n+1}\subseteq V_{n}$ for each $n$. For each $k\in\mathbb{N}$, the convergence of $\{x_{n}\}_{n=1}^{\infty}$ says $V_{k}\cap\{x_{n}\}_{n=1}^{\infty}\neq\emptyset$. Therefore, we take $n_{k}$ satisfying $n_{k}\geq k$ and $x_{n_{k}}\in V_{k}$. We need to clam that the subsequence $\{x_{n_{k}}\}_{k=1}^{\infty}$ converges to $x$. Given any neighborhood $N_{x}$ of $x$, there exists $k_{0}$ satisfying $V_{k_{0}}\subseteq N_{x}$. For $k\geq k_{0}$, we have
\[x_{n_{k}}\in V_{k}\subseteq V_{k_{0}}\subseteq N_{x},\]
which says that the subsequence $\{x_{n_{k}}\}_{k=1}^{\infty}$ converges to $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top31}}\tag{52}\mbox{}\end{equation}
Proposition \ref{top31}. Let $\{x_{\gamma}\}_{\gamma\in\Gamma}$ be a net, and let ${\cal A}$ be a family of sets such that the following conditions are satisfied.

The net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in each member of ${\cal A}$.
The intersection of any two members of ${\cal A}$ contains a member of ${\cal A}$.

Then, there is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$ which is eventually in each member of ${\cal A}$.

Proof. Since the intersection of any two members of ${\cal A}$ contains a member of ${\cal A}$, it says that the family ${\cal A}$ is directed by $\subseteq$.
We define a set $\Gamma$ be the set of all pairs $(\gamma ,A)$ given below
\[\Gamma =\left\{(\gamma ,A):\gamma\in\Gamma, A\in {\cal A}\mbox{ and }x_{\gamma}\in A\right\}.\]
Let $(\gamma_{1},A_{1})$ and $(\gamma_{2},A_{2})$ be any two members of $\Gamma$. Then, there exist $A\in {\cal A}$ and $\gamma^{*}\in\Gamma$ satisfying $A\subseteq (A_{1}\cap A_{2})$, $\gamma^{*}\succ\gamma_{1}$ and $\gamma^{*}\succ\gamma_{2}$.
The first condition says that there exists $\gamma\in\Gamma$ satisfying $\gamma\succ\gamma^{*}$ and $x_{\gamma}\in A$. In this case, we also have $\gamma\succ\gamma_{1}$ and $\gamma\succ\gamma_{2}$ by the transitivity. In other words, there exists $(\gamma ,A)\in\Gamma$ satisfying $(\gamma ,A)\succ(\gamma_{1},A_{1})$ and $(\gamma ,A)\succ(\gamma_{2},A_{2})$. This shows that $\Gamma$ is directed by the product ordering for $\Gamma\times {\cal A}$.

We define a function $N:\Gamma\rightarrow\Gamma$ by $N(\gamma ,A)=\gamma$. If $(\gamma_{1},A_{1})\succ(\gamma_{2},A_{2})$ then
$N(\gamma_{1},A_{1})=\gamma_{1}\succ\gamma_{2}=N(\gamma_{2},A_{2})$. The first condition says that the range of $N$ is confinal in $\Gamma$
Proposition \ref{top30} says that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet of $\{x_{\gamma}\}_{\gamma\in\Gamma}$. Finally, given any $A\in {\cal A}$ and any $\gamma\in\Gamma$ with $x_{\gamma}\in A$, i.e., $(\gamma ,A)\in\Gamma$, if $(\gamma^{*},A^{*})\in\Gamma$ with
$(\gamma^{*},A^{*})\succ (\gamma ,A)$, then
\[x_{N(\gamma^{*},A^{*})}=x_{\gamma^{*}}\in A^{*}\subseteq A.\]
This shows that the subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is eventually in $A$, and the proof is complete. $\blacksquare$

By definition, we see that a point $x$ is a cluster point of a net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ if and only if $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in every neighborhood of $x$.

A net may have one, many, or no cluster points.
If a net converges to a point, then this point is surely a cluster point.
It is possible that a net may have a single cluster point and fail to converge to this point.

\begin{equation}{\label{top33}}\tag{53}\mbox{}\end{equation}
Proposition \ref{top33}. Let $\{x_{\gamma}\}_{\gamma\in\Gamma}$ be a net in a topological space. For each $\gamma_{0}\in\Gamma$, let $A_{\gamma_{0}}$ be the set of all points $x_{\gamma}$ for $\gamma\succ\gamma_{0}$. Then $x$ is a cluster point of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ if and only if $x\in\mbox{\em cl}(A_{\gamma})$ for each $\gamma\in\Gamma$.

Proof. If $x$ is a cluster point of $\{x_{\gamma}\}_{\gamma\in\Gamma}$, then, for each $\gamma\in\Gamma$, $A_{\gamma}$ intersects each neighborhood of $s$, since $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in each neighborhood of $x$, i.e., $x\in\mbox{cl}(A_{\gamma})$ for each $\gamma\in\Gamma$. For the converse, if $x$ is not a cluster point of $\{x_{\gamma}\}_{\gamma\in\Gamma}$, then there exists a neighborhood $N_{x}$ of $x$ such that $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is not frequently in $N_{x}$. Therefore, for some $\gamma_{0}\in\Gamma$,
if $\gamma\succ\gamma_{0}$, then $x_{\gamma}\not\in N_{x}$, i.e., $N_{x}\cap A_{\gamma_{0}}=\emptyset$, which says that $x\not\in\mbox{cl}(A_{\gamma_{0}})$. This contradiction completes the proof. $\blacksquare$

Recall that the first axiom of countability says that the neighborhood system of each point has a countable base. In other words, for each point $x\in X$,
there is a countable family of neighborhoods of $x$ such that every neighborhoods of $x$ contains some member of the family.

\begin{equation}{\label{top34}}\tag{54}\mbox{}\end{equation}
Proposition \ref{top34}. Let $X$ be a topological space satisfying the first axiom of countability. Then, we have the following properties.

(i) A point $x$ is an accumulation point of a set $A$ if and only if there exists a sequence in $A\setminus\{s\}$ which converges to $x$.

(ii) If $x$ is a cluster point of a sequence $\{x_{n}\}_{n=1}^{\infty}$, then there exists a subsequence of $\{x_{n}\}_{n=1}^{\infty}$ converging to $x$.

Proof. To prove part (i), suppose that $x$ is an accumulation point of a subset $A$ of $X$, and that $\{N_{x}^{(n)}\}_{n=1}^{\infty}$ is a countable base for the neighborhood system of $x$. Let $M_{x}^{(n)}=\bigcap_{i=1}^{n}N_{x}^{(i)}$. Then $\{M_{x}^{(n)}\}_{n=1}^{\infty}$ is also a countable base for the neighborhood system of $x$ with $M_{x}^{(n+1)}\subseteq M_{x}^{(n)}$. For each $n\in\mathbb{N}$, we select a point $x_{n}$ from $M_{x}^{(n)}\cap (A\setminus\{x\})\neq\emptyset$. Then, we obtain a sequence $\{x_{n}\}_{n=1}^{\infty}$ which evidently converges to $x$. The converse is obvious.

To prove part (ii), suppose that $x$ is a cluster point of a sequence $S$, and that $\{M_{x}^{(n)}\}_{n=1}^{\infty}$ is a countable base for the neighborhood system of $x$ satisfying $M_{x}^{(n+1)}\subseteq M_{x}^{(n)}$ for each $n\in\mathbb{N}$, which is constructed from part (i). For each $i\in\mathbb{N}$, we can choose $n_{i}$ satisfying $n_{i}\geq i$ and $x_{n_{i}}\in M_{x}^{(i)}$. Then $\{x_{n_{i}}\}_{i\in\mathbb{N}}$ is a subsequence of $\{x_{n}\}_{n=1}^{\infty}$ which converges to $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top40}}\tag{55}\mbox{}\end{equation}
Proposition \ref{top40}. Let $f$ be a function from the topological space $(X,\tau_{X})$ into the topological space $(Y,\tau_{Y})$. Then $f$ is continuous on $X$ if and only if, for every net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ that converges to $x$, the net $\{f(x_{\gamma})\}_{\gamma\in\Gamma}$ also converges to $f(x)$.

Proof. Suppose that $f$ is continuous, and that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. Now we assume that the net $\{f(x_{\gamma})\}_{\gamma\in\Gamma}$ does not converge to $f(x)$. Proposition \ref{top35} says that there exists a subnet $\{f(x_{N_{\beta}})\}_{\beta\in\Gamma}$ of $\{f(x_{\gamma})\}_{\gamma\in\Gamma}$ such that no subnet of $\{f(x_{N_{\beta}})\}_{\beta\in\Gamma}$ converges to $f(x)$; that is, the subnet $\{f(x_{N_{\beta}})\}_{\beta\in\Gamma}$ does not converge to $f(x)$.
Therefore, there exists a neighborhood $N$ of $f(x)$ such that there does not exist any $\beta_{0}\in\Gamma$ such that $\beta\succ\beta_{0}$ implies $f(x_{N_{\beta}})\in N$. Since $f$ is continuous, it means that $f^{-1}(N)$ is a neighborhood of $x$. Therefore, there exists a neighborhood $f^{-1}(N)$ of $x$ such that there does not exist any $\beta_{0}\in\Gamma$ such that $\beta\succ\beta_{0}$ implies $x_{N_{\beta}}\in f^{-1}(N)$; that is, the subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ does not converge to $x$. Since the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$, the contradiction occurs by part (ii) of Proposition \ref{top36}.

For the converse, suppose that, for every net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ that converges to $x$, the net $\{f(x_{\gamma})\}_{\gamma\in\Gamma}$ also converges to $f(x)$. We also assume that $f$ is not continuous. Then, using Proposition \ref{top39},
there exists a neighborhood $V$ of $f(x)$ such that no neighborhood $N_{x}$ of $x$ satisfies $f(N_{x})\subseteq V$; that is, each neighborhood $N_{x}$ of $x$ satisfies $f(N_{x})\not\subseteq V$. We consider the directed set $\Gamma\equiv {\cal N}_{x}$ consisting of all neighborhoods of $x$. We form a net $\{y_{\gamma}\}_{\gamma\in\Gamma}$ by taking $y_{\gamma}$ satisfying $y_{\gamma}\in N_{x}\equiv\gamma$ and $f(y_{\gamma})\not\in V$. Given any neighborhood $N_{x}^{(0)}$ of $x$, we take $\gamma_{0}\equiv N_{x}^{(0)}$. For $\gamma\equiv N_{x}\succ\gamma_{0}\equiv N_{x}^{(0)}$, i.e., $N_{x}\subseteq N_{x}^{(0)}$, we have $x\in N_{x}\subseteq N_{x}^{(0)}$, which shows that the net $\{y_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. However the net $\{f(y_{\gamma})\}_{\gamma\in\Gamma}$ does not converge to $f(x)$, since
$f(y_{\gamma})\not\in V$ for all $\gamma\in\Gamma$. This completes the proof. $\blacksquare$

Suppose that $\{{\bf x}_{\gamma}\}_{\gamma\in\Gamma}$ is a net in the countable product space $(\prod_{i\in I}X_{i},\tau )$. The $j$th-coordinate of ${\bf x}_{\gamma}$ is denoted by $x_{\gamma}^{(j)}$. This says that $\{x_{\gamma}^{(j)}\}_{\gamma\in\Gamma}$ is a net in $X_{j}$. Then, we have the following interesting result.

\begin{equation}{\label{top64}}\tag{56}\mbox{}\end{equation}
Proposition \ref{top64}. Suppose that $\{{\bf x}_{\gamma}\}_{\gamma\in\Gamma}$ is a net in the product space $(\prod_{i\in I}X_{i},\tau )$. Then, the net $\{{\bf x}_{\gamma}\}_{\gamma\in\Gamma}$ converges to ${\bf x}$ if and only if the net $\{x_{\gamma}^{(j)}\}_{\gamma\in\Gamma}$ converges to $x^{(j)}$, where $x^{(j)}$ is the $j$th-coordinate of ${\bf x}$.

Proof. Suppose that the net $\{{\bf x}_{\gamma}\}_{\gamma\in\Gamma}$ converges to ${\bf x}$. The projection mapping $p_{j}:\prod_{i\in I}X_{i}\rightarrow X_{j}$ is continuous. Proposition \ref{top40} says that the net $\{p_{j}({\bf x}_{\gamma}\}_{\gamma\in\Gamma}=\{x_{\gamma}^{(j)})\}_{\gamma\in\Gamma}$ converges to $p_{j}({\bf x})=x^{(j)}$. For the converse, let $U=\prod_{i\in I}U_{i}$ be a neighborhood of ${\bf x}$ in the product topology, where each $U_{j}$ is a neighborhood of $x^{(j)}$ satisfying $U_{j}=X_{j}$ except finitely many $j_{1},\cdots ,j_{n}$. For each $j\in I$ except $j_{1},\cdots ,j_{n}$, we have $x_{\gamma}^{(j)}\in U_{j}=X_{j}$ for any $\gamma\in\Gamma$. For $j_{1},\cdots ,j_{n}$, since net $\{x_{\gamma}^{(j)}\}_{\gamma\in\Gamma}$ converges to $x^{(j)}$, we can find $\gamma_{1},\cdots ,\gamma_{n}$ such that if $\gamma_{k}\prec\gamma$ then $x_{\gamma}^{(j_{k})}\in U_{j_{k}}$ for $k=1,\cdots ,n$. Since $\Gamma$ is directed, there exists $\gamma_{0}\in\Gamma$ such that if $\gamma_{0}\prec\gamma$ then $x_{\gamma}^{(j_{k})}\in U_{j_{k}}$ for $k=1,\cdots ,n$. Combining these two facts, it follows that if $\gamma_{0}\prec\gamma$ then $x_{\gamma}^{(j)}\in U_{j}$ for each $j\in I$, i.e., ${\bf x}_{\gamma}\in U=\prod_{i\in I}U_{i}$. This shows that net $\{{\bf x}_{\gamma}\}_{\gamma\in\Gamma}$ converges to ${\bf x}$, and the proof is complete. $\blacksquare$

Let $(D,\succ_{D})$ and $(E,\succ_{E})$ be two directed sets. The Cartesian product $D\times E$ is directed by $\succ$ that is defined by
\[(d_{1},e_{1})\succ (d_{2},e_{2})\mbox{ if and only if }d_{1}\succ_{D}d_{2}\mbox{ and }e_{1}\succ_{E}e_{2}.\]
The directed set $(D\times E,\succ )$ is the product directed set. We also want to define the product of a family of directed sets. Suppose that we are given a family of directed sets $\{(D_{\lambda},\succ_{\lambda})\}_{\lambda\in\Gamma}$. The Cartesian product $\prod_{\lambda\in\Gamma}D_{\lambda}$ is the set of all functions $d$ defined on $\Gamma$ such that $d_{\lambda}=d(\lambda )$ is a member of $D_{\lambda}$ for each $\lambda\in\Gamma$.
For $d,e\in\prod_{\lambda\in\Gamma}D_{\lambda}$, we define the product order $\succ$ on $\prod_{\lambda\in\Gamma}D_{\lambda}$ by
\[d\succ e\mbox{ if and only if }d_{\lambda}\succ_{\lambda}e_{\lambda}\mbox{ for each }\lambda\in\Gamma .\]
We can verify that the product directed set $(\prod_{\lambda\in\Gamma}D_{\lambda},\succ )$ is indeed a directed set.

An important special case of the product directed set is that in which all coordinate sets $D_{\lambda}$ are identical to $D$ and all orderings $\succ_{\lambda}$ are identical to $\succ^{*}$. In this case, $\prod_{\lambda\in\Gamma}D$ is simply the set $D^{\Gamma}$ of all functions defined on $\Gamma$ to $D$, which is directed by the convention that $d\succ e$ if and only if $d_{\lambda}\succ^{*}e_{\lambda}$ for each $\lambda\in\Gamma$.

Consider the class of all functions $S$ such that $S(m,n)$ is defined whenever $m$ belongs to a directed set $D$ and $n$ belongs to a directed set $E_{m}$. We want to find a net $R$ with values in this domain such that $S\circ R$ converges to $\lim_{m}\lim_{n} S(m,n)$ whenever $S$ is a function to a topological space and this iterated limit exists. It is interesting to notice that the solution of this problem requires Moore-Smith convergence, for, considering double sequences, no sequence whose range is a subset of $\omega\times\omega$ can have this property, where $\omega$ is the set of nonnegative integers. The construction which yields a solution to the problem is a variant of the diagonal process. Let $F$ be the product directed set $D\times (\prod_{m\in D}E_{m})$, and for each point $(m,f)$ of $F$ let $R(m,f)=(m,f(m))$. Then $R$ is the required net.

Proposition. Let $D$ be a directed set, let $E_{m}$ be a directed set for each $m$ in $D$, let $F$ be the product $D\times (\prod_{m\in D}E_{m})$, and for $(m,f)$ in $F$ let $R(m,f)=(m,f(m))$. If $S(m,n)$ is a member of a topological space for each $m$ in $D$ and each $n$ in $E_{m}$, then $S\circ R$ converges to $\lim_{m}\lim_{n}S(m,n)$ whenever this iterated limit exists.

Proof. Suppose that $\lim_{m}\lim_{n}S(m,n)=s$, and that $U$ is an open neighborhood of $s$. We must find a member $(m,f)$ of $F$ such that, if $(p,g)\succ (m,f)$, then $S\circ R(p,g)\in U$. Choose $m\in D$ so that $\lim_{n}S(p,n)\in U$ for each $p$ following $m$ and then, for each such $p$, choose a member $f(p)$ of $E_{p}$ such that $S(p,n)\in U$ for all $n$ following $f(p)$ in $E_{p}$. If $p$ is a member of $D$ which does not follow $m$ let $f(p)$ be an arbitrary member of $E_{p}$. If $(p,g)\succ (m,f)$, then $p\geq m$, hence $\lim_{n}S(p,n)\in U$ and, since $g(p)\geq f(p)$, $S\circ R(p,g)=S(p,g(p))\in U$. $\blacksquare$

Suppose that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in $A$. Then, the set $\Lambda$ of all members $\gamma\in\Gamma$ satisfying $x_{\gamma}\in A$ has the following property:
\begin{equation}{\label{top27}}\tag{57}
\mbox{for each $\gamma\in\Gamma$ there exists $\gamma_{0}\in\Lambda$ satisfying $\gamma_{0}\succ\gamma$.}
\end{equation}
The subset of $\Gamma$ satisfying (\ref{top27}) is called cofinal.

Each cofinal subset $\Lambda$ of a directed set $(\Gamma,\prec)$ is also directed by $\prec$, since, for elements $\alpha$ and $\beta$ in $\Lambda$, there exists $\gamma\in\Gamma$ satisfying $\gamma\succ\alpha$ and $\gamma\succ\beta$. Condition (\ref{top27}) says that there exists $\gamma_{0}\in\Lambda$ satisfying $\gamma_{0}\succ\gamma$, which also implies $\gamma_{0}\succ\alpha$ and $\gamma_{0}\succ\beta$.
We have the following obvious equivalence: a net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is frequently in a set $A$ if and only if a cofinal subset of $\Gamma$ maps into the set $A$, and this is the case if and only if the net is not eventually in the complement of $A$.

\begin{equation}{\label{l}}\tag{L}\mbox{}\end{equation}

Filters.

There is an alternative approach to the concept of convergence in a general topological space through the notion of a filter. We would expect that, since nets and filters are both designed for the study of convergence, there will have to be many theorems about filters completely analogous to theorems stated in terms of nets.

Definition. Let $X$ be any set. A collection ${\cal F}$ of nonempty subsets of $X$ is said to be a filter on $X$ when the following conditions are satisfied:

${\cal F}\neq\emptyset$;
if $A,B\in {\cal F}$ then $A\cap B\in {\cal F}$;
if $A\in {\cal F}$ and $A\subseteq B$ then $B\in {\cal F}$. $\sharp$

Example. We have many examples for filters.

(i) From Proposition \ref{top8}, we see that the family ${\cal N}_{x}$ of all neighborhoods of $x$ is a filter on $X$.

(ii) Let $X$ be a set, and let $Y$ be a nonempty subset of $X$. Then, the family ${\cal F}$ of all subsets of $X$ which contains $Y$ is a filter on $X$. More precisely, the following family
\[{\cal F}=\left\{F\subseteq X:Y\subseteq F\right\}\]
is a filter on $X$.

(iii) Let $X$ be any set, and let ${\cal D}$ be a nonempty family of subsets of $X$ such that if $D_{1},D_{2}\in {\cal D}$ then there exists $D_{3}\in {\cal D}$ satisfying $D_{3}\subseteq D_{1}\cap D_{2}$. Let
\[{\cal F}=\left\{A:B\subseteq A\mbox{ for some }B\in{\cal D}\right\}.\]
Then ${\cal F}$ is a filter on $X$ (Exercise!). In this case, the family ${\cal F}$ is said to be a filter generated by ${\cal D}$, and ${\cal D}$ is said to be a base for the filter ${\cal F}$.

(iv) Let $\{x_{\alpha}\}_{\alpha\in\Lambda}$ be a net in $X$, and let ${\cal D}$ be the family of all subsets of the form
\begin{equation}{\label{top43}}\tag{58}
D_{\beta}=\left\{x_{\alpha}:\beta\prec\alpha\right\}
\end{equation}
for all $\beta\in\Lambda$. If $D_{\beta_{1}},D_{\beta_{2}}\in {\cal D}$ then there exists $D_{\beta_{3}}\in {\cal D}$ satisfying $D_{\beta_{3}}\subseteq D_{\beta_{1}}\cap D_{\beta_{2}}$ (Exercise!). Part (iii) says that ${\cal D}$ forms a base for a filter ${\cal F}$ given by
\[{\cal F}=\left\{A:D_{\beta}\subseteq A\mbox{ for some }\beta\in\Lambda\right\}.\]
In this case, the family ${\cal F}$ is said to be the filter generated by the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$. $\sharp$

Definition. Let $(X,\tau )$ be a topological space, and let ${\cal F}$ be a filter on $X$.

${\cal F}$ is said to converge to $x$ if and only if every neighborhood of $x$ is a members of ${\cal F}$.
${\cal F}$ is said to have $x$ as a limit point if and only if every neighborhood of $x$ meets every member of ${\cal F}$. $\sharp$

Proposition. Let $\{x_{\alpha}\}_{\alpha\in\Lambda}$ be a net in a topological space $(X,\tau )$, and let ${\cal F}$ be the filter generated by the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$. Then, we have the following properties.

(i) The filter ${\cal F}$ converges to $x$ if and only if the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ converges to $x$.

(ii) The filter ${\cal F}$ has a limit point if and only if $x$ is a limit point of the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$.

Proof. To prove part (i), suppose that the filter ${\cal F}$ converges to $x$. Then given any neighborhood $N_{x}$ of $x$, $N_{x}\in {\cal F}$; that is, there exists $\beta\in\Lambda$ satisfying $D_{\beta}\subseteq N_{x}$. This also says that if $\alpha\succ\beta$ then $x_{\alpha}\in D_{\beta}\subseteq N_{x}$. Therefore, the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ converges to $x$. For the converse, suppose that the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ converges to $x$. Then, given any neighborhood $N_{x}$ of $x$, there exists $\beta\in\Lambda$ such that $\alpha\succ\beta$ (i.e., $x_{\alpha}\in D_{\beta})$ implies $x_{\alpha}\in N_{x}$, which says $D_{\beta}\subseteq N_{x}$. This also says $N_{x}\in {\cal F}$. Therefore, the filter ${\cal F}$ converges to $x$. Part (ii) is left as an exercise. $\blacksquare$

Let ${\cal F}$ be a filter on a set $X$. We shall construct a net from ${\cal F}$. For any $A,B\in {\cal F}$, we write $A\prec B$ if and only if $B\subseteq A$.
It is easy to verify that this ordering makes ${\cal F}$ as a directed set. For each $A\in {\cal F}$, we can select an element $x_{A}\in A$. Then, we can form a net $\{x_{A}\}_{A\in {\cal F}}$, which is called a {\bf net induced by the filter ${\cal F}$}.

Proposition. The filter ${\cal F}$ converges to $x$ if and only if each net $\{x_{A}\}_{A\in {\cal F}}$ induced by the filter ${\cal F}$ converges to $x$.

Proof. Suppose that the filter ${\cal F}$ converges to $x$. Let $N_{x}$ be any neighborhood of $x$. Then $N_{x}\in {\cal F}$. If $A\in {\cal F}$ and $A\succ N_{x}$, i.e., $A\subseteq N_{x}$, then $x_{A}\in A\subseteq N_{x}$. This shows that $\{x_{A}\}_{A\in {\cal F}}$ converges to $x$. For the converse, suppose that the filter ${\cal F}$ does not converge to $x$. Then, there exists a neighborhood of $N_{x}$ of $x$ with $N_{x}\not\in {\cal F}$. For any $A\in {\cal F}$, if $A\setminus N_{x}=\emptyset$ then $A\subseteq N_{x}$, which contradicts $N_{x}\not\in {\cal F}$. Therefore, we must have $A\setminus N_{x}\neq\emptyset$. Then, we can select an element $x_{A}\in A\setminus N_{x}$ and form a net $\{x_{A}\}_{A\in {\cal F}}$ that is induced by ${\cal F}$ and does not converge to $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top42}}\tag{59}\mbox{}\end{equation}
Proposition \ref{top42}. Let ${\cal F}$ be a filter on a topological space $(X,\tau )$. Then $x$ is a limit point of ${\cal F}$ if and only if there exists a filter ${\cal F}^{*}$ such that ${\cal F}\subseteq {\cal F}^{*}$ and ${\cal F}^{*}$ converges to $x$.

Proof. Suppose that $x$ is a limit point of ${\cal F}$. Let
\[{\cal D}^{*}=\left\{A\cap N_{x}:A\in {\cal F}\mbox{ and $N_{x}$ is a neighborhood of $x$}\right\}.\]
Then ${\cal D}^{*}$ is a base that generates the filter ${\cal F}^{*}$ given by
\[{\cal F}^{*}=\left\{A:B\subseteq A\mbox{ for some }B\in {\cal D}^{*}\right\}.\]
Since $A\cap N_{x}\subseteq N_{x}$, it says that $N_{x}\in {\cal F}^{*}$; that is, every neighborhood $N_{x}$ of $x$ is a member of ${\cal F}^{*}$, which says that the filter ${\cal F}^{*}$ converges to $x$. We remain to show ${\cal F}\subseteq {\cal F}^{*}$. Since $X$ is a neighborhood of $x$, for each $A\in {\cal F}$, we have $A=A\cap X\in {\cal D}^{*}\subseteq {\cal F}^{*}$.

For the converse, let $N_{x}$ be any neighborhood of $x$ and let $A\in {\cal F}$. Then $N_{x},A\in {\cal F}^{*}$, i.e., $\emptyset\neq A\cap N_{x}\in {\cal F}^{*}$ by definition. In other words, given any neighborhood $N_{x}$ of $x$ and any $A\in{\cal F}$, since $A\cap N_{x}\neq\emptyset$, it says that $x$ is a limit point of ${\cal F}$. This completes the proof. $\blacksquare$

Proposition. Suppose that $f$ is a function from a topological space $(X,\tau_{X})$ into a topological space $(Y,\tau_{Y})$. Let ${\cal F}_{X}$ be a filter on $X$. Then, we have the following properties.

(i) Let
\[f\left ({\cal F}_{X}\right )=\left\{f(A):A\in {\cal F}_{X}\right\}.\]
Then $f({\cal F}_{X})$ is the base for some filter ${\cal F}_{Y}$ on $Y$.

(ii) $f$ is continuous if and only if given any filter ${\cal F}_{X}$ on $X$ such that the filter ${\cal F}_{X}$ converges to $X$, the filter ${\cal F}_{Y}$ for which $f({\cal F}_{X})$ is the base for ${\cal F}_{Y}$ converges to $f(x)$. $\sharp$

Proposition. Let $A$ be a subset of a topological space $(X,\tau )$. Then $x\in\mbox{cl}(A)$ if and only if there exists a filter ${\cal F}$ on $X$ such that $A\in {\cal F}$ and the filter ${\cal F}$ converges to $x$. $\sharp$

Proposition. The topological space $(X,\tau )$ is a $T_{2}$-space if and only if any convergent filter on $X$ converges to a unique limit. $\sharp$

Definition. An ultrafilter on a set $X$ is a maximal filter on $X$; that is, a filer ${\cal F}$ on a set $X$ is an ultrafilter if and only if, given any filter ${\cal F}^{*}$ finer than ${\cal F}$, i.e., ${\cal F}\subseteq {\cal F}^{*}$, we have ${\cal F}={\cal F}^{*}$. $\sharp$

Example. Let $X$ be any set, and let $x\in X$. Then, the family of all subsets containing $x$ forms an ultrafilter ${\cal F}$ on $X$. Let ${\cal F}^{*}$ be any filter finer than ${\cal F}$. Given any $A^{*}\in {\cal F}^{*}$, we consider the following cases.

If $x\in A^{*}$ then $A^{*}\in {\cal F}$.
If $x\not\in A^{*}$ then $x\in X\setminus A^{*}$, which says that $X\setminus A^{*}\in {\cal F}\subseteq {\cal F}^{*}$. The definition of filter says
\[\emptyset =A^{*}\cap X\setminus A^{*}\in {\cal F}^{*},\]
which contradicts the fact that each member of ${\cal F}^{*}$ is nonempty.

The above two cases say that $x$ must be in each member of ${\cal F}^{*}$, i.e., ${\cal F}^{*}\subseteq {\cal F}$. This claims that ${\cal F}$ is indeed an ultrafilter. $\sharp$

\begin{equation}{\label{top41}}\tag{60}\mbox{}\end{equation}
Proposition \ref{top41}. Let ${\cal F}$ be a filter on a set $X$. Then ${\cal F}$ is an ultrafilter on $X$ if and only if given any subset $A$ of $X$, either $A\in {\cal F}$ or $X\setminus A\in {\cal F}$.

Proof. ${\cal F}^{*}$ is a filter on $X$ with ${\cal F}\subseteq {\cal F}^{*}$. Now, given any $A^{*}\in {\cal F}^{*}$, we assume $A^{*}\not\in {\cal F}$. Then, we have
\[X\setminus A^{*}\in {\cal F}\subseteq {\cal F}^{*}.\]
By the definition of filter, we must have
\[\emptyset =A^{*}\cal (X\setminus A^{*})\in {\cal F}^{*},\]
which contradicts that each member of ${\cal F}^{*}$ is nonempty. Therefore, we must have $A^{*}\in {\cal F}$, which shows that ${\cal F}$ is an ultrafilter on $X$. Suppose now that ${\cal F}$ is an ultrafilter on $X$. We also assume that there exists a subset $A$ of $X$ such that neither $A$ nor $X\setminus A$ is in ${\cal F}$. We want to find a filter ${\cal F}^{*}$ which is strictly finer than ${\cal F}$. We consider the following cases.

Suppose that $A\cap B\neq\emptyset$ for each $B\in {\cal F}$. Then, we can take
\[{\cal D}=\left\{A\cap B:B\in {\cal F}\right\}\]
as a filter base generating a filter ${\cal F}^{*}$ which is strictly finer than ${\cal F}$.
Suppose that $A\cap B^{\circ}=\emptyset$ for some $B^{\circ}\in {\cal F}$. We also assume that $(X\setminus A)\cap B_{0}=\emptyset$ for some $B_{0}\in {\cal F}$. Then $B^{\circ}\cap B_{0}\in {\cal F}$ and
\[B^{\circ}\cap B_{0}=\left [(B^{\circ}\cap B_{0})\cap A\right ]\cup\left [(B^{\circ}\cap B_{0})\cap (X\setminus A)\right ]
\subseteq (B^{*}\cap A)\cup\left [(X\setminus A)\cap B_{0}\right ]=\emptyset .\]
This contradiction says that $(X\setminus A)\cap B\neq\emptyset$ for each $B\in {\cal F}$. Therefore, we can take
\[{\cal D}=\left\{(X\setminus A)\cap B:B\in {\cal F}\right\}\]
as a filter base generating a filter ${\cal F}^{*}$ which is strictly finer than ${\cal F}$.

This completes the proof. $\blacksquare$

\begin{equation}{\label{top44}}\tag{61}\mbox{}\end{equation}
Proposition \ref{top44}. Let $X$ be any set. Then every filter on $X$ is contained in an ultrafilter.

Proof. Let $\mathfrak{U}$ be the family of all filters ${\cal F}^{*}$ on $X$ satisfying ${\cal F}\subseteq {\cal F}^{*}$. Then, the family $\mathfrak{U}$ can ne partially ordered by “is finer than”. Suppose that $\mathfrak{C}=\{{\cal F}^{*}_{\gamma}\}_{\gamma\in\Gamma}$ is a chain in $\mathfrak{U}$. We want to claim that
\[{\cal D}=\left\{B:B\in {\cal F}^{*}_{\gamma}\mbox{ for some }\gamma\in\Gamma\right\}\]
is a filter base generating a filter ${\cal F}^{\circ}$. For $B_{0},B_{1}\in {\cal D}$, we see that $B_{0}\in {\cal F}^{*}_{\gamma_{0}}$ and $B_{1}\in {\cal F}^{*}_{\gamma_{1}}$ for some $\gamma_{0},\gamma_{1}\in\Gamma$. Since $\mathfrak{C}$ is a chain, we have either ${\cal F}^{*}_{\gamma_{0}}\subseteq {\cal F}^{*}_{\gamma_{1}}$ or ${\cal F}^{*}_{\gamma_{1}}\subseteq {\cal F}^{*}_{\gamma_{0}}$. Suppose that ${\cal F}^{*}_{\gamma_{0}}\subseteq {\cal F}^{*}_{\gamma_{1}}$. Then $B_{0},B_{1}\in {\cal F}^{*}_{\gamma_{1}}$, i.e., $B_{0}\cap B_{1}\in {\cal F}^{*}_{\gamma_{1}}$, which says that $B_{0}\cap B_{1}\in {\cal D}$. Therefore ${\cal D}$ is indeed a filter base generating a filter ${\cal F}^{\circ}$. It is clearly that ${\cal F}^{\circ}$ is finer than any member of $\mathfrak{C}$, i.e., ${\cal F}^{\circ}$ is an upper bound in $\mathfrak{C}$. In other words, each chain in $\mathfrak{U}$ has an upper bound. Applying the Zorn’s lemma, the family contains a maximal element which is an ultrafilter containing ${\cal F}$. This completes the proof. $\blacksquare$

Proposition. Let $f$ be a function from a set $X$ onto another set $Y$, let ${\cal F}_{X}$ be an ultrafilter on $X$, and let $f\left ({\cal F}_{X}\right )$
be the filter base generating the filter ${\cal F}_{Y}$ on $Y$. Then ${\cal F}_{Y}$ is an ultrafilter on $Y$.

Proof. Given $A\subseteq Y$, in order to show that ${\cal F}_{Y}$ is an ultrafilter, we must claim either $A\in {\cal F}_{Y}$ or $X\setminus A\in {\cal F}_{Y}$ by Proposition \ref{top41}. Since ${\cal F}_{X}$ is an ultrafilter, using Proposition~\ref{top41} again, we have either $f^{-1}(A)\in {\cal F}$ or $f^{-1}(Y\setminus A)=X\setminus f^{-1}(A)\in {\cal F}$. If $f^{-1}(A)\in {\cal F}$ then
\[f(f^{-1}(A))=A\in f({\cal F}_{X})\subseteq {\cal F}_{Y}.\]
If $X\setminus f^{-1}(A)\in {\cal F}_{X}$ then $Y\setminus A\in {\cal F}_{Y}$. This shows that ${\cal F}_{Y}$ is an ultrafilter o $Y$, and the proof is complete. $\blacksquare$

Proposition. Let ${\cal F}$ be an ultrafilter on a topological space $(X,\tau )$. If $x$ is a limit point of ${\cal F}$, then the filter ${\cal F}$ converges to $x$.

Proof. Since $x$ is a limit point, Proposition~\ref{top42} says that there exists a filter ${\cal F}^{*}$ finer than ${\cal F}$ which converges to $x$. Since ${\cal F}$ is an ultrafilter, we have ${\cal F}^{*}={\cal F}$. This completes the proof. $\blacksquare$

The concept of finer filter is a kind of concept of subnet. Inspired by Proposition \ref{top41}, we have the following definition.

Definition. The net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ in a set $X$ is said to be an ultranet when, given any subset $A$ of $X$, there exists $\alpha_{0}\in\Lambda$ satisfying either $x_{\alpha}\in A$ for all $\alpha\succ\alpha_{0}$ or $x_{\alpha}\in X\setminus A$ for all $\alpha\succ\alpha_{0}$. $\sharp$

Example. Let $X$ be any set and let $x\in X$. The net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ defined by $x_{\alpha}=x$ for all $\alpha\in\Lambda$ is an ultranet, since either $x\in A$ or $x\in X\setminus A$ for any given subset $A$ of $X$. $\sharp$

Proposition. We have the following properties.

(i) Let ${\cal F}$ be the filter generated by the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$. Then $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet if and only if ${\cal F}$ is an ultrafilter.

(ii) If ${\cal F}$ is an ultrafilter then every net induced by the filter ${\cal F}$ is an ultranet.
\end{enumerate}

Proof. To prove part (i), suppose that $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet. Let ${\cal D}$ be a filter based defined in (\ref{top42}). Given any subset $A$ of $X$, since $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet, there exists $\beta_{0}\in\Lambda$ satisfying $D_{\beta_{0}}\subseteq A$ or $D_{\beta_{0}}\subseteq X\setminus A$. Therefore, we have either $A\in {\cal F}$ or $X\setminus A\in {\cal F}$, which says that ${\cal F}$ is an ultrafilter by Proposition \ref{top41}. Suppose that ${\cal F}$ is an ultrafilter. Given any subset $A$ of $X$, we have either $A\in {\cal F}$ or $X\setminus A\in {\cal F}$. According to the construction of the filter ${\cal F}$, there exists $\alpha_{0}\in\Lambda$ satisfying $D_{\alpha_{0}}\subseteq A$ or $D_{\alpha_{0}}\subseteq X\setminus A$, which says that either $x_{\alpha}\in A$ for all $\alpha\succ\alpha_{0}$ or $x_{\alpha}\in X\setminus A$ for all $\alpha\succ\alpha_{0}$. This says that $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet. Part (ii) is left as an exercise. $\blacksquare$

\begin{equation}{\label{top62}}\tag{62}\mbox{}\end{equation}
Proposition \ref{top62}. Let $f$ be a function from a set $X$ onto a set $Y$. If $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet in $X$, then $\{f(x_{\alpha})\}_{\alpha\in\Lambda}$ is an ultranet in $Y$.

Proof. Given any subset $A$ of $Y$, since $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet in $X$, there exists $\alpha_{0}\in\Lambda$ satisfying either $x_{\alpha}\in f^{-1}(A)$ for all $\alpha\succ\alpha_{0}$ or $x_{\alpha}\in X\setminus f^{-1}(A)$ for all $\alpha\succ\alpha_{0}$. Therefore, there exists $\alpha_{0}\in\Lambda$ satisfying either $f(x_{\alpha})\in A$ for all $\alpha\succ\alpha_{0}$ or $f(x_{\alpha})\in Y\setminus A$ for all $\alpha\succ\alpha_{0}$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top56}}\tag{63}\mbox{}\end{equation}
Proposition \ref{top56}. Given any net $\{x_{\alpha}\}_{\alpha\in\Lambda}$, there exists a subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of $\{x_{\alpha}\}_{\alpha\in\Lambda}$ such that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is an ultranet.

Proof. Let ${\cal F}$ be the filter generated by the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$. Proposition \ref{top44} says that ${\cal F}$ is contained in an ultrafilter ${\cal F}^{*}$. Given any $A\in {\cal A}^{*}$, suppose that there exists $\alpha_{0}\in\Lambda$ satisfying $x_{\alpha}\in X\setminus A$ for $\alpha\succ\alpha_{0}$. Then, we have
\[X\setminus A\in {\cal F}\subseteq {\cal F}^{*}.\]
Therefore, we also have
\[\emptyset =A\cap (X\setminus A)\in {\cal F}^{*}.\]
This contradiction says that, given any $\alpha_{0}\in\Lambda$, there exists $\alpha\in\Lambda$ satisfying $\alpha\succ\alpha_{0}$ and $x_{\alpha}\in A$. Let
\[\Gamma =\left\{(\alpha ,A):\mbox{$\alpha\in\Lambda$ and $A\in {\cal F}^{*}$ with $x_{\alpha}\in A$}\right\}.\]
Since ${\cal F}^{*}$ is directed by $A_{1}\prec A_{2}$ if $A_{2}\subseteq A_{1}$, it follows that $\Gamma$ is a directed set. Now we define $N:\Gamma\rightarrow\Lambda$ by $N(\alpha ,A)=\alpha$. Then $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is a subnet. Given any $A\in {\cal F}^{*}$, the definition of $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ also says that there exists $\beta_{0}\in\Gamma$ such that $x_{N_{\beta}}\in A$ for $\beta\succ\beta_{0}$. Since ${\cal F}^{*}$ is an ultrafilter, it contains any subset $A$ of $X$ or its complement $X\setminus A$. Therefore, there exists $\beta_{0}\in\Gamma$ satisfying $x_{N_{\beta}}\in A$ for $\beta\succ\beta_{0}$ or $x_{N_{\beta}}\in X\setminus A$ for $\beta\succ\beta_{0}$. This shows that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is an ultranet, and the proof is complete. $\blacjsquare$

\begin{equation}{\label{top55}}\tag{64}\mbox{}\end{equation}
Proposition \ref{top55}. Let $\{x_{\alpha}\}_{\alpha\in\Lambda}$ be an ultranet in a topological space $(X,\tau )$. If $x$ is a cluster point of $\{x_{\alpha}\}_{\alpha\in\Lambda}$, then the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ converges to $x$.

Proof. Let $N_{x}$ be any neighborhood of $x$. Since $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is an ultranet. There exists $\alpha_{0}\in\Lambda$ satisfying either $x_{\alpha}\in N_{x}$ for all $\alpha\succ\alpha_{0}$ or $x_{\alpha}\in X\setminus N_{x}$ for all $\alpha\succ\alpha_{0}$.
Since $x$ is a cluster point of $\{x_{\alpha}\}_{\alpha\in\Lambda}$, we cannot have the case of $x_{\alpha}\in X\setminus N_{x}$ for all $\alpha\succ\alpha_{0}$. In other words, we must have $x_{\alpha}\in N_{x}$ for all $\alpha\succ\alpha_{0}$, which says that the net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ converges to $x$. This completes the proof. $\blacksquare$

\begin{equation}{\label{m}}\tag{M}\mbox{}\end{equation}

Product Spaces.

Given sets $X_{1},X_{2},\cdots ,X_{n}$, the Cartesian product of these sets is defined by
\[\prod_{i=1}^{n}X_{i}=\left\{\left (x_{1},x_{2},\cdots ,x_{n}\right ):x_{i}\in X_{i}\mbox{ for }i=1,\cdots ,n\right\}.\]
Let $I$ be a countable index set. Given $X_{i}$ for $i\in I$, we can also define the product by
\[\prod_{i\in I}X_{i}=\left\{{\bf x}=\left (x_{1},x_{2},\cdots ,x_{n},\cdots\right ):x_{i}\in X_{i}\mbox{ for }i\in I\right\}.\]
The function
\[p_{j}:\prod_{i\in I}X_{i}\rightarrow X_{j}\]
is called the projection into the $i$th component.

Let $(X,\tau_{X})$ and $(Y,\tau_{Y})$ be two topological spaces, and let ${\cal B}$ be the family of all Cartesian products $O_{X}\times O_{Y}$, where $O_{X}$ is a $\tau_{X}$-open subset of $X$ and $O_{Y}$ is a $\tau_{Y}$-open subset of $Y$. Since
\[\left (O_{X}^{(1)}\times O_{Y}^{(1)}\right )\cap \left (O_{X}^{(2)}\times O_{Y}^{(2)}\right )
=\left (O_{X}^{(1)}\cap O_{X}^{(2)}\right )\times \left (O_{Y}^{(1)}\cap O_{Y}^{(2)}\right ),\]
i.e., the intersection of two members of ${\cal B}$ is a member of ${\cal B}$. Corollary \ref{top11} says that ${\cal B}$ is a base for some topological space $(X\times Y,\tau_{X\times Y})$. The formal definition is given below.

\begin{equation}{\label{top126}}\tag{65}\mbox{}\end{equation}
Definition \ref{top126}. Let $(X,\tau_{X})$ and $(Y,\tau_{Y})$ be two topological spaces. We define a topology on the product set $X\times Y$ by taking the following base
\[{\cal B}=\left\{O_{X}\times O_{Y}:O_{X}\in\tau_{X}\mbox{ and }O_{Y}\in\tau_{Y}\right\},\]
which is called the product topology for $X\times Y$. The spaces $X$ and $Y$ are called the coordinate spaces. $\sharp$

The functions $P_{X}$ and $P_{Y}$ denote the projections into the coordinate spaces $X$ and $Y$, respectively. Suppose that $X\times Y$ is endowed with the product topology. Then, we have the following observations.

A subset $W$ of $X\times Y$ is open regarding the product topology if and only if, for each element $(x,y)$ of $W$, there exist $\tau_{X}$-neighborhood $N_{x}$ of $x$ and $\tau_{Y}$-neighborhood $N_{y}$ of $y$ satisfying $N_{x}\times N_{y}\subseteq W$.
The projection $P_{X}$ is a $\tau_{X}$-continuous function, since if $O_{X}$ is $\tau_{X}$-open then $P_{X}^{-1}(O_{X})=O_{X}\times Y$ is open with respect to the product topology. Also, the projection $P_{Y}$ is $\tau_{Y}$-continuous.
Suppose $\tau$ is a topology for $X\times Y$ such that each of the projections is continuous. For $O_{X}\in\tau_{X}$ and $O_{Y}\in\tau_{Y}$, since
\[O_{X}\times O_{Y}=\left (O_{X}\times Y\right )\cap\left (X\times O_{Y}\right )
=P_{X}^{-1}\left (O_{X}\right )\cap P_{Y}^{-1}\left (O_{Y}\right ),\]
the continuity of the projections says that the set $O_{X}\times O_{Y}$ is $\tau$-open. This also says that $\tau$ is finer than the product topology. In other words, the product topology is the coarsest topology such that the projections into coordinate spaces are continuous.

There is no difficulty for extending the definition of product topology to Cartesian product of any finite number of coordinate spaces. Let $X_{1},X_{2},\cdots ,X_{n}$ be topological spaces. Then, a base for the product topology is the family of all products $O_{1}\times\cdots\times O_{n}$, where each $O_{i}$ is open in $X_{i}$.

Example. Let $(X,d_{X})$ and $(Y,d_{Y})$ be two metric spaces. We can form a new metric space called the Cartesian product $X\times Y$ whose set
of points is the set
\[X\times Y\equiv\{(x,y):x\in X, y\in Y\}\]
and whose metric $d_{X\times Y}$ is defined by
\[d_{X\times Y}\left ((x_{1},y_{1}),(x_{2},y_{2})\right )=\sqrt{d_{X}(x_{1},x_{2})^{2}+d_{Y}(y_{1},y_{2})^{2}}.\]
Then, the product topology agrees with the topology induced by the product metric $d_{X\times Y}$. $\sharp$

Proposition. Let ${\bf x}=(x_{1},\cdots ,x_{n})$ and ${\bf y}=(y_{1},\cdots ,y_{n})$ be any two points in $\mathbb{R}^{n}$. Define
\[d({\bf x},{\bf y})=\max_{i=1,\cdots ,n}\left |x_{i}-y_{i}\right |.\]
Then $(\mathbb{R}^{n},d)$ is a metric space. Moreover, the topology on $\mathbb{R}^{n}$ induced by the metric $d$ is the same as the product topology on $\mathbb{R}^{n}$.

Proof. The proof of $(\mathbb{R}^{n},d)$ being a metric space is left as an exercise. Let $\tau$ be the product topology on $\mathbb{R}^{n}$, and let $\tau_{d}$ be the induced topology by the metric $d$. We are going to use Proposition \ref{top67} to prove $\tau =\tau_{d}$. The neighborhood system ${\cal N}_{\bf x}$ of ${\bf x}$ for $\tau$ is given by
\[{\cal N}_{\bf x}=\left\{\prod_{i=1}^{n}\left (x_{i}-r_{i},x_{i}+r_{i}\right ):r_{i}>0\mbox{ and }i=1,\cdots ,n\right\},\]
where $(x_{i}-r_{i},x_{i}+r_{i})$ denotes an open interval centered at $x_{i}$ with radius $r_{i}$. The neighborhood system ${\cal N}_{\bf x}^{(d)}$ of ${\bf x}$ for $\tau_{d}$ is given by
\[{\cal N}_{\bf x}^{(d)}=\left\{B({\bf x};r):r>0\right\}.\]
For ${\bf y}\in B({\bf x};r)$, it means
\[\max_{i=1,\cdots ,n}\left |x_{i}-y_{i}\right |=d({\bf x},{\bf y})<r,\]
which implies $|x_{i}-y_{i}|<r$ for all $i=1,\cdots ,n$. This says that $B({\bf x};r)\in {\cal N}_{\bf x}$. For
\[N_{x}=\prod_{i=1}^{n}\left (x_{i}-r_{i},x_{i}+r_{i}\right )\in {\cal N}_{\bf x},\]
let $r=\min\{r_{1},\cdots ,r_{n}\}$. Then $B({\bf x};r)\in {\cal N}_{\bf x}^{(d)}$ and $B({\bf x};r)\subseteq N_{x}$. Using Proposition \ref{top67}, the proof is complete. $\blacksquare$

Definition. Let $\{(X_{i},\tau_{i})\}_{i\in I}$ be a countable family of topological spaces. Define the following family
\[{\cal S}=\left\{\prod_{i\in I}O_{i}:O_{i}=X_{i}\mbox{ for all but at most one $i$ and $O_{i}\in\tau_{i}$ for all $i\in I$}\right\}.\]
The topology $\tau$ on $\prod_{i\in I}X_{i}$ such that ${\cal S}$ is a subbase is called the product topology. $\sharp$

Proposition. Given the product space $(\prod_{i\in I}X_{i},\tau )$, each projection
\[p_{j}:\prod_{i\in I}X_{i}\rightarrow X_{j}\]
is continuous. Moreover, the product topology $\tau$ is the coarsest topology on $\prod_{i\in I}X_{i}$ such that each $p_{j}$ is continuous.

Proof. It is obvious that the product topology defined in this way makes each projection $p_{j}$ to be continuous. If any topology on $\prod_{i\in I}X_{i}$ were strictly coarser than the product topology, then some member of the subbase ${\cal S}$ would not be open, which says that at least one of the projection could not be continuous. This completes the proof. $\blacksquare$

\begin{equation}{\label{top15}}\tag{66}\mbox{}\end{equation}
Proposition \ref{top15}. Given the product space $(\prod_{i\in I}X_{i},\tau )$, define the following family
\[{\cal B}=\left\{\prod_{i\in I}O_{i}:O_{i}=X_{i}\mbox{ for all but at most finitely many $i$ and $O_{i}\in\tau_{i}$ for all $i\in I$}\right\}.\]
Then ${\cal B}$ is a base for the product topology.

Proof. We see that a base for $\tau$ is obtained by taking all finite intersections of members of ${\cal S}$. Let $S_{1},\cdots ,S_{n}$ be members of ${\cal S}$, where $S_{k}=\prod_{i\in I}O_{i}$with $O_{i}=X_{i}$ for all $i$ except possibly for $i=i_{k}$ and $k=1,\cdots ,n$. Then, we have
\[\bigcap_{k=1}^{n}S_{k}=\prod_{i\in I}O_{i},\]
where $O_{i}=X_{i}$ for all $i$ except possibly for $i_{1},i_{2},\cdots ,i_{n}$. This means that each member of the base induced from the subbase ${\cal S}$ is a member of ${\cal B}$. On the other hand, each member of ${\cal B}$ is the intersection of finitely many members of ${\cal S}$. This shows that ${\cal B}$ is indeed a base for the product topology $\tau$, and the proof is complete. $\blacksquare$

Proposition. Given the product space $(\prod_{i\in I}X_{i},\tau )$, each $(X_{i},\tau_{i})$ is homeomorphic to a subspace of $(\prod_{i\in I}X_{i},\tau )$ given by
\[Y_{i}=\left\{\left (y_{1},y_{2},\cdots ,y_{i-1},x,y_{i+1},\cdots ,y_{n},\cdots\right ):x\in X_{i}\right\}\]
for each $i\in I$, where each $y_{j}$ is fixed for $j\neq i$.

Proof. It suffices to prove the case of $(X_{1},\tau_{1})$, since the same argument could be used for the cases of any $i\in I$. Let $y_{i}\in X_{i}$ for $i\geq 2$ be fixed. We define the function $q_{1}$ from $X_{1}$ into $\prod_{i\in I}X_{i}$by
\[q_{1}(x)=(x,y_{2},\cdots ,y_{n},\cdots )\mbox{ for each }x\in X_{1}.\]
Let $Y$ be the subspace of $(\prod_{i\in I}X_{i},\tau )$ defined by
\[Y_{1}=\left\{\left (x,y_{2},\cdots ,y_{n},\cdots\right ):x\in X_{1}\right\}.\]
Then $q_{1}:S_{1}\rightarrow Y$ is one-to-one and onto. We remain to show that $q_{1}$ and $q_{1}^{-1}$ are continuous. Given any $O\in\tau_{Y_{1}}$. Then $O=Y\cap O^{*}$ for some $O^{*}\in\tau$. If $p_{1}$ is the projection into the first component $X_{1}$, then $p_{1}(O^{*})\in\tau_{1}$. We have $q_{1}^{-1}(O)=p_{1}(O^{*})$, which says $q_{1}^{-1}(O)\in\tau_{1}$. This shows that $q_{1}$ is continuous. On the other hand, given any $O^{\circ}\in\tau_{1}$, we have
\[q_{1}(O^{\circ})=(q_{1}^{-1})^{-1}(O^{\circ})=Y\cap\prod_{i\in I}O_{i},\]
where $O_{i}=X_{i}$ for $i\geq 2$ and $Q_{1}=O^{\circ}$. It follows $(q_{1}^{-1})^{-1}(O^{\circ})\in\tau_{Y}$. This shows that $q_{1}^{-1}$ is continuous. Therefore $q_{1}$ is a homeomorphism. This completes the proof. $\blacksquare$

\begin{equation}{\label{top17}}\tag{67}\mbox{}\end{equation}
Proposition \ref{top17}. Let $f$ be a function from the topological space $(X,\tau_{X})$ into the product space $(\prod_{i\in I}X_{i},\tau )$. For each $j\in I$, we define the function
\[f_{j}:(X,\tau_{X})\rightarrow (X_{j},\tau_{j})\mbox{ by }f_{j}(x)=(p_{j}\circ f)(x)\]
for each $x\in X$, where each $p_{j}$ is the projection into the $j$th component. Then $f$ is continuous if and only if each $f_{j}$ is continuous.

Proof. Let $f$ be continuous. Since each $f_{j}$ is the composition of two continuous functions, it shows that $f_{j}$ is also continuous. For the converse, suppose that $\prod_{i\in I}O_{i}$ is any member of the base ${\cal B}$ given in Proposition \ref{top15}, where $O_{i}=X_{i}$ for each $i\in I$ except for $i_{1},\cdots ,i_{m}$. For every $i\in I$ except for $i_{1},\cdots ,i_{m}$, since $O_{i}=X_{i}$, we have $f^{-1}(O_{i})=X$. Note that $f^{-1}(\prod_{i\in I}O_{i})$ is the set of all points $x\in X$ satisfying $f(x)\in\prod_{i\in I}O_{i}$. Then, we have
\[f^{-1}\left (\prod_{i\in I}O_{i}\right )=\bigcap_{i\in I}f_{i}^{-1}(O_{i})
=f_{i_{1}}^{-1}(O_{i_{1}})\cap\cdots\cap f_{i_{m}}^{-1}(O_{i_{m}}).\]
Since $f_{i}$ is continuous, i.e., $f_{i_{k}}^{-1}(O_{i_{k}})\in\tau_{X}$ for $k=1,\cdots ,m$, it follows $f^{-1}(\prod_{i\in I}O_{i})\in\tau_{X}$. Proposition \ref{top16} says that $f$ is continuous. This completes the proof. $\blacksquare$

Example. Let $\mathbb{R}$ be endowed with the usual topology induced by the absolute value metric, and let $\mathbb{R}^{2}$ be endowed with the product topology from $\mathbb{R}$. For the function $f:\mathbb{R}\rightarrow\mathbb{R}^{2}$ defined by
\[f(x)=(\cos x,e^{x})\mbox{ for each }x\in\mathbb{R}.\]
Then $f_{1}(x)=\cos x$ and $f_{2}(x)=e^{x}$. Since $f_{1}$ and $f_{2}$ are continuous, Proposition \ref{top17} says that $f$ is continuous. $\sharp$

\begin{equation}{\label{top20}}\tag{68}\mbox{}\end{equation}
Proposition \ref{top20}. The product space $(\prod_{i\in I}X_{i},\tau )$ is a $T_{1}$-space if and only if each $(X_{i},\tau_{i})$ is a $T_{1}$-space. $\sharp$

\begin{equation}{\label{top23}}\tag{69}\mbox{}\end{equation}
Proposition \ref{top23}. Suppose that the product space $(\prod_{i\in I}X_{i},\tau )$ is a $T_{1}$-space. Then, each $(X_{i},\tau_{i})$ is homeomorphic to a closed subspace of the product space $(\prod_{i\in I}X_{i},\tau )$. $\sharp$

Proposition. The product space $(\prod_{i\in I}X_{i},\tau )$ is a $T_{2}$-space if and only if each $(X_{i},\tau_{i})$ is a $T_{2}$-space.

Proof. Assume that each $(X_{i},\tau_{i})$ is $T_{2}$-space. Given any two distinct points ${\bf x}$ and ${\bf y}$ in $\prod_{i\in I}X_{i}$, we shall use $x_{i}$ and $y_{i}$ to denote the $i$th component of ${\bf x}$ and ${\bf y}$, respectively. Since ${\bf x}\neq {\bf y}$, we must have that $x_{i^{*}}\neq y_{i^{*}}$ for some $i^{*}\in $latex . Therefore, there exist $O_{i^{*}}^{(1)},O_{i^{*}}^{(2)}\in\tau_{i}$ such that $x_{i^{*}}\in O_{i^{*}}^{(1)}$, $y_{i^{*}}\in O_{i^{*}}^{(2)}$ and $O_{i^{*}}^{(1)}\cap O_{i^{*}}^{(2)}=\emptyset$. Let $O^{(1)}=\prod_{i\in I}H_{i}$ with $H_{i}=X_{i}$ for $i\neq i^{*}$ and $H_{i^{*}}=O_{i^{*}}^{(1)}$, and let $O^{(2)}=\prod_{i\in I}G_{i}$ with $G_{i}=X_{i}$ for $i\neq i^{*}$ and $G_{i^{*}}=O_{i^{*}}^{(2)}$. Then $O^{(1)}$ and $O^{(2)}$ are the neighborhoods of ${\bf x}$ and ${\bf y}$, respectively. Since any point of $O^{(1)}$ differs from any point of$O^{(2)}$ at least in the $i^{*}$th component, it follows that $O^{(1)}\cap O^{(2)}=\emptyset$. This completes the proof. $\blacksquare$

Proposition. The product space $(\prod_{i\in I}X_{i},\tau )$ is a $T_{3}$-space if and only if each $(X_{i},\tau_{i})$ is a $T_{3}$-space, and the product space $(\prod_{i\in I}X_{i},\tau )$ is a regular space if and only if each $(X_{i},\tau_{i})$ is a regular space.

Proof. Suppose that the product space $(\prod_{i\in I}X_{i},\tau )$ is $T_{3}$-space. Since each $(X_{i},\tau_{i})$ is homeomorphic to a subspace of the product space $(\prod_{i\in I}X_{i},\tau )$, using Proposition~\ref{top21}, it follows that each $(X_{i},\tau_{i})$ is a $T_{3}$-space. For the converse, given any ${\bf x}\in O_{\bf x}$ that is open in the product space, there exists a member $B_{\bf x}$ of the base ${\cal B}$ for the product topology satisfying ${\bf x}\in B_{\bf x}\subseteq O_{\bf x}$, where $B_{\bf x}$ is also open in the product space. Then, we see that $B_{\bf x}=\prod_{i\in I}B_{i}$, where $B_{i}=X_{i}$ for all $i$ except for $i_{1},\cdots ,i_{m}$ and $B_{i}\in\tau_{i}$ for all $i$. Since each $X_{i}$ is $T_{3}$-space, Proposition \ref{top22} says that there exist open sets $O_{i_{k}}^{*}\in\tau_{i_{k}}$ satisfying
\[x_{i_{k}}\in O_{i_{k}}^{*}\subset\mbox{cl}(O_{i_{k}}^{*})\subseteq B_{i_{k}}\]
for $k=1,\cdots ,m$, where $x_{i}$ denotes the $i$th component of ${\bf x}$. We define $O_{\bf x}^{*}=\prod_{i\in I}O_{i}^{*}$ with $O_{i}^{*}=X_{i}$ for all $i\neq i_{k}$ for $k=1,\cdots ,m$. Then, we obtain
\[{\bf x}\in O_{\bf x}^{*}\subset\mbox{cl}(O_{\bf x}^{*})=\mbox{cl}\left (\prod_{i\in I}O_{i}^{*}\right )\subset\prod_{i\in I}\mbox{cl}\left (O_{i}^{*}\right )\subset\prod_{i\in I}B_{i}=B_{\bf x}\subseteq O_{\bf x}.\]
Proposition \ref{top21} says that the product space is a $T_{3}$-space. Finally, using Proposition~\ref{top20}, we see that the product space $(\prod_{i\in I}X_{i},\tau )$ is a regular space if and only if each $(X_{i},\tau_{i})$ is a regular space. This completes the proof. $\blacksquare$

\begin{equation}{\label{top24}}\tag{70}\mbox{}\end{equation}
Proposition \ref{top24}. Suppose that the product space $(\prod_{i\in I}X_{i},\tau )$ is a normal space. Then, each $(X_{i},\tau_{i})$ is a normal space.

Proof. Proposition \ref{top23} says that each $(X_{i},\tau_{i})$ is homeomorphic to a closed subspace of $(\prod_{i\in I}X_{i},\tau )$. Therefore Proposition \ref{top25} says that each $(X_{i},\tau_{i})$ is a normal space. $\blacksquare$

We need to remark that the converse of Proposition \ref{top24} is not true.

Proposition. The product space of a countable family of nonempty topological space is separable if and only if each component space is separable. $\sharp$

Proposition. The product space of a countable family of nonempty topological space satisfies the second axiom of countability if and only if each component space satisfies the second axiom of countability.

Proof. Let $\{(X_{n},\tau_{n})\}_{n=1}^{\infty}$ be a countable family of topological space satisfies the second axiom of countability, and let ${\cal B}_{n}$ be the countable base for $\tau_{n}$. Define
\[{\cal B}=\left\{\prod_{n=1}^{\infty}V_{n}:V_{n}=X_{n}\mbox{ for all but at most finitely many
$n$ and $V_{n}\in {\cal B}_{n}$ for $V_{n}\neq X_{n}$}\right\}.\]
It is easily to verify that ${\cal B}$ is a countable base for the product topology on the product space $\prod_{n=1}^{\infty}X_{n}$; that is, $\prod_{n=1}^{\infty}X_{n}$ satisfies the second axiom of countability. On the other hand, suppose that each $X_{n}$ satisfies the second axiom of countability. Since each component space is homeomorphic to a subspace of the product space, using Remark \ref{top47}, it follows that each component space also satisfies the second axiom of countability. This completes the proof. $\blacksquare$

In general, given an (uncountable) family of nonempty sets $\{X_{\alpha}\}_{\alpha\in\Lambda}$, the Cartesian product
\[X\equiv\prod_{\alpha\in\Lambda}X_{\alpha}\]
is defined to be the set of all functions $x\in X$ on $\Lambda$ satisfying $x_{\alpha}=x(\alpha )\in X_{\alpha}$ for each $\alpha\in\Lambda$. The set $X_{\alpha}$ is the $\alpha$-th coordinate set. The projection $P_{\alpha}$ of the product set into the $\alpha$-th coordinate set is defined by $P_{\alpha}(x)=x_{\alpha}$. Suppose that $\tau_{\alpha}$ is a topology for the coordinate set $X_{\alpha}$. The product topology will make each projection $P_{\alpha}$ to be continuous. It is necessary and sufficient that each set of the form
\begin{equation}{\label{top215}}\tag{71}
P_{\alpha}^{-1}(O_{\alpha})=\left\{\prod_{\beta\in\Lambda}O_{\beta}:O_{\alpha}\in\tau_{\alpha}
\mbox{ and }O_{\beta}=X_{\beta}\mbox{ for }\beta\neq\alpha\right\}
=\left\{x\in X:x_{\alpha}=x(\alpha )\in O_{\alpha}\right\}
\end{equation}
is open with respect to the product topology, where $O_{\alpha}$ is a $\tau_{\alpha}$-open subset of $X_{\alpha}$. The product topology is defined by taking the family of all sets in the form of (\ref{top215}) as the subbase. It is clear that the product topology is the coarsest topology such that all the projections are continuous. A base for the product topology is the family of all finite intersections of subbase elements. The formal definition is given below.

Definition. Let $\{(X_{\alpha },\tau_{\alpha })\}_{\alpha\in\Lambda}$ be a family of topological spaces. The product topology on $\prod_{\alpha\in\Lambda}X_{\alpha}$ is defined by taking the following base
\[{\cal B}=\left\{\prod_{\alpha\in\Lambda}O_{\alpha}:O_{\alpha}\in\tau_{\alpha}
\mbox{ and }O_{\alpha}=X_{\alpha}\mbox{ except for a finite number of $\alpha$}\right\}.\]

It is not true that $\prod_{\alpha\in\Lambda}U_{\alpha}$ is always open with respect to the product topology when each $U_{\alpha}$ is $\tau_{\alpha}$-open for each $\alpha\in\Lambda$.

Definition. A function $f$ from a topological space $(X,\tau_{X})$ into another space $(Y,\tau_{Y})$ is said to be open when the image of each $\tau_{X}$-open set is $\tau_{Y}$-open; that is, if $O_{X}$ is $\tau_{X}$-open, then $f(O_{X})$ is $\tau_{Y}$-open. $\sharp$

\begin{equation}{\label{top216}}\tag{72}\mbox{}\end{equation}
Proposition \ref{top216}. The projection of a product topological space into each of its coordinate spaces is open. In other words, the projection of an open set with respect to the product topology is open.

Proof. Let $P_{c}$ be the projection of $\prod_{a\in A}X_{a}$ into $X_{c}$. In order to show that $P_{c}$ is open, it is sufficient to show that the image
of a neighborhood of a point $x$ in the product is a neighborhood of $P_{c}(x)$, and it may be assumed that the neighborhood in the product space is a member of the defining base for the product topology. Suppose that $x\in V=\{y:y_{a}\in U_{a},a\in F\}$, where $F$ is a finite subset of $A$ and $U_{a}$ is open in $X_{a}$ for each $a\in F$. We construct a copy of $X_{c}$ which contains the point $x$. For $z\in X_{c}$, let $(f(z))_{c}=z$, and for $a\neq c$, let $(f(z))_{a}=x_{a}$. Then $P_{c}\circ f(z)=z$. If $c\not\in F$, then clearly $f(X_{c})\subseteq V$ and $P_{c}(V)=X_{c}$ which is open. If $c\in F$, then $f(z)\in V$ if and only if $z\in U_{c}$ and $P_{c}(V)=U_{c}$. The result follows. (As a matter of fact, the function $f$ defined in this proof is a homeomorphism). $\blacksquare$

We remark that the projection of a closed set with respect to the product topology is not necessarily closed.

\begin{equation}{\label{top217}}\tag{73}\mbox{}\end{equation}
Proposition \ref{top217}. A function $f$ from a topological space into a product topological space $\prod_{\alpha\in\Lambda}X_{\alpha}$ is continuous if and only if the composition $P_{\alpha}\circ f$ is continuous for each projection $P_{\alpha}$.

Proof. If $f$ is continuous, then $P_{\alpha}\circ f$ is always continuous since $P_{\alpha}$ is continuous. If $P_{\alpha}\circ f$ is continuous for each $\alpha$, then for each open subset $O_{\alpha}$ of $X_{\alpha}$, the set
\[\left (P_{\alpha}\circ f\right )^{-1}\left (O_{\alpha}\right )=f^{-1}\left (P_{\alpha}^{-1}\left (O_{\alpha}\right )\right )\]
is open. It follows that the inverse image under $f$ of each member of the defining subbase for the product topology is open, and hence $f$ is continuous by part (iii) of Proposition \ref{top218}. $\blacksquare$

\begin{equation}{\label{top219}}\tag{74}\mbox{}\end{equation}
Proposition \ref{top219}. A net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ in a product topological space converges to $x$ if and only if its projection in each coordinate space converges to the projection of $x$.

Proof. Since the projection into each coordinate space is continuous, the net $\{P_{\alpha}(x_{\gamma})\}_{\gamma\in\Gamma}$ surely converges to $P_{\alpha}(x)$. For the converse, since the net $\{P_{\alpha}(x_{\gamma})\}_{\gamma\in\Gamma}$ converges to $P_{\alpha}(x)$ for each $\gamma\in\Gamma$, for each open set $O_{\alpha}$ containing $P_{\alpha}(x)$, we see that $\{P_{\alpha}(x_{\gamma})\}_{\gamma\in\Gamma}$ is eventually in $O_{\alpha}$, which says that $\{x_{\gamma}\}_{\gamma\in\Gamma}$ is eventually in $P_{\alpha}^{-1}(O_{\alpha})$. Therefore $\{x_{\gamma}\}_{\gamma\in\Gamma}$ must eventually be in each finite intersection of sets of the form $P_{\alpha}^{-1}(O_{\alpha})$. Since the family of such finite intersections is a base for the neighborhood system of $x$ with respect to the product topology, it follows that the net $\{x_{\gamma}\}_{\gamma\in\Gamma}$ converges to $x$. This completes the proof. $\blacksquare$

Convergence in the product topology is called coordinatewise convergence. In particular, if all the coordinate spaces are identical, then the convergence in the product topology is also called pointwise convergence. In this special case, the Cartesian product $\prod_{\alpha\in\Lambda}X$ is simply the set of all functions from $\Lambda$ to $X$ and is denoted by $X^{\Lambda}$.

\begin{equation}{\label{top220}}\tag{75}\mbox{}\end{equation}
Remark \reftop220}. Using Proposition\ref{top219}, it follows that a net $\{f_{\gamma}\}_{\gamma\in\Gamma}$ in $X^{\Lambda}$ converges to $f$ with respect to the topology of pointwise convergence (i.e. the product topology) if and only if the net $\{f_{\gamma}(\alpha )\}_{\gamma\in\Gamma}$
converges to $f(\alpha )$ for each $\alpha\in\Lambda$. $\sharp$

Proposition. The product of Hausdorff spaces is a Hausdorff space.

Proof. If $x$ and $y$ are distinct members of the product $\prod_{\alpha\in\Lambda}X_{\alpha}$, then $x_{\alpha}\neq y_{\alpha}$ for some $\alpha\in\Lambda$. If each coordinate space is Hausdorff, then there are disjoint open neighborhoods $U$ and $V$ of $x_{\alpha}$ and $y_{\alpha}$ respectively and $P_{\alpha}^{-1}(U)$ and $P_{\alpha}^{-1}(V)$ are disjoint neighborhoods of $x$ and $y$ in the product. $\blacksquare$

Proposition. Let $\{(X_{\alpha},\tau_{\alpha})\}_{\alpha\in\Lambda}$ be a family of topological spaces satisfying the first axiom of countability. Then, the product topological space $\prod_{\alpha\in\Lambda}X_{\alpha}$ satisfies the first axiom of countability if and only if all but a countable number of the spaces $X_{\alpha}$ are indiscrete. $\sharp$

Proposition. The product space $\prod_{i=1}^{\infty}X_{i}$ of the countable family of nonempty topological spaces $\{(X_{i},\tau_{i})\}_{i=1}^{\infty}$ is connected if and only if each component space $X_{i}$ is connected.

Proof. Suppose that $\prod_{i=1}^{\infty}X_{i}$ is connected. Since the projection $p_{j}$ from $\prod_{i=1}^{\infty}X_{i}$ into $X_{j}$ is continuous, it follows that each $X_{j}$ is connected. $\blacksquare$

For the converse, suppose that $\prod_{i=1}^{\infty}X_{i}$ is disconnected. Then $\prod_{i=1}^{\infty}X_{i}=S\cup T$ can be expressed as the union of nonempty disjoint open sets $S$ and $T$ in $\prod_{i=1}^{\infty}X_{i}$. Given any ${\bf u}\in S$ and ${\bf v}\in T$, since $S$ is open, there exists a neighborhood $\prod_{i=1}^{\infty}N_{i}$ of ${\bf u}$ such that $\prod_{i=1}^{\infty}N_{i}\subseteq S$, where $N_{i}$ is $\tau_{i}$-open and $N_{i}=X_{i}$ except for finite numbers $i_{1},\cdots ,i_{n}$. Define $c_{i}^{(0)}=v_{i}$ for $i\neq i_{1},\cdots ,i_{n}$ and $c_{i}^{(0)}=u_{i}$ for $i=i_{1},\cdots ,i_{n}$. Then
\[{\bf c}^{(0)}=\left (c_{1}^{(0)},\cdots ,c_{n}^{(0)},\cdots\right )\in\prod_{i=1}^{\infty}N_{i}\subseteq S.\]
Define ${\bf c}^{(1)}$ by $c_{i}^{(1)}=v_{i}$ for $i\neq i_{1},\cdots ,i_{n}$, $c_{i}^{(1)}=v_{i}$ for $i=i_{1}$ and $c_{i}^{(1)}=u_{i}$ for $i=i_{2},\cdots ,i_{n}$. Inductively, we define ${\bf c}^{(m)}$ by setting $c_{i}^{(m)}=v_{i}$ for $i\neq i_{m+1},\cdots ,i_{n}$ and $c_{i}^{(m)}=u_{i}$ for $i=i_{m+1},\cdots ,i_{n}$. Then ${\bf c}^{(n)}={\bf v}$. Let
\[A_{m}=\left\{{\bf x}:\mbox{$x_{i_{m}}$ is arbitrary, $x_{i_{m+1}}=u_{i_{m+1}},\cdots ,
x_{i_{n}}=u_{i_{n}}$ and otherwise $x_{i}=v_{i}$}\right\}.\]
Then ${\bf c}^{(m-1})\in A_{m-1}\cap A_{m}$ for $1\leq m\leq n$ and ${\bf v}={\bf c}^{(n)}\in A_{n}$. We also see that $A_{1},\cdots ,A_{n}$ form a chain from $S$ to $T$. We want to claim that each $A_{i}$ is connected. Define a function $g_{m}$ from $X_{i_{m}}$ onto $A_{m}$ as follows.

$g_{m}(x_{i_{m}})$ is the point of $A_{m}$ such that the $i_{m}$th coordinate is $x_{i_{m}}$.
The $i_{k}$th coordinate is $u_{i_{k}}$ for $k=m+1,\cdots ,n$.
All the other $i$th coordinate is $v_{i}$.

Then each $g_{m}$ is continuous. Since the inverse of $g_{m}$ is the projection from $A_{m}$ onto $X_{i_{m}}$, it says that $g_{m}$ is a homeomorphism. Since each $X_{i_{m}}$ is connected, it follows that each $A_{m}$ is connected; that is to say, the union $\bigcup_{i=1}^{m}A_{i}$ is connected. Since $\bigcup_{i=1}^{m}A_{i}$ meets both of $S$ and $T$, a contradiction occurs by Proposition~\ref{top88}, and the proof is complete. $\blacksquare$

General Topology.

Closed Sets.

Interior and Boundary.

Bases and Subbases.

Subspaces.

Neighborhood System.

Continuity.

Quotient Space and Quotient Topology.

Connected Sets.

The Separation Axioms.

Convergence.

Filters.

Product Spaces.

Hsien-Chung Wu

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General Topology.

Closed Sets.

Interior and Boundary.

Bases and Subbases.

Subspaces.

Neighborhood System.

Continuity.

Quotient Space and Quotient Topology.

Connected Sets.

The Separation Axioms.

Convergence.

Filters.

Product Spaces.

Hsien-Chung Wu

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