Compact Spaces

Charles Burton Barber (1845-1894) was an English painter.

We have the sections

Compactness
Equivalent Compactness in Metric Spaces
Compactification

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

A family ${\cal A}=\{A_{\gamma}\}_{\gamma\in\Gamma}$ is a {\bf cover} of $A$ when $A$ is a subset of the union $\bigcup_{\gamma\in\Gamma}A_{\gamma}$. The family is an open cover of $A$ when each member of ${\cal A}$ is an open set. A subcover of $A$ is a subfamily of ${\cal A}$ which is also a cover of $A$.
The topological space $(X,\tau )$ is said to be a Lindelöf space when every open cover of $X$ has a countable open subcover. $\sharp$

\begin{equation}{\label{top50}}\tag{1}\mbox{}\end{equation}
Proposition \ref{top50}. Let $(X,\tau )$ be a topological space having a countable base, and let $A$ be a subset of $X$. Then, each open cover of $A$ has a countable open subcover of $A$. In other words, any topological space satisfying the second axiom of countability is a Lindel\”{o}f space.

Proof. Let ${\cal B}=\{B_{n}\}_{n=1}^{\infty}$ be a countable base for the topological space $(X,\tau )$. Suppose that ${\cal U}=\{U_{\gamma}\}_{\gamma\in\Gamma}$ is any open cover of $A$. Since each open set $U_{\gamma}$ is the union of members of ${\cal B}$ by Proposition 9 in page topological spaces, there is a subfamily ${\cal C}$ of ${\cal B}$ such that ${\cal C}$ also covers $A$ and, for each $C_{i}\in {\cal C}$, there exists $U_{\gamma_{i}}\in {\cal U}$ satisfying $C_{i}\subseteq U_{\gamma_{i}}$. Since ${\cal C}$ covers $A$, it follows that $\{U_{\gamma_{i}}\}_{i=1}^{\infty}$ is an open subcover of $A$. This completes the proof. $\blacksquare$

\begin{equation}{\label{top49}}\tag{2}\mbox{}\end{equation}
Proposition \ref{top49}. Every closed subspace of a Lindelöf space is also a Lindelöf space.

Proof. Let $(X,\tau_{X})$ be a Lindelöf space, and let $(Y,\tau_{Y})$ be a closed subspace of $(X,\tau_{X})$, which means that $Y$ is a $\tau_{X}$-closed subset of $X$. Let $\{U_{\gamma}\}_{\gamma\in\Gamma}$ be a family of $\tau_{Y}$-open cover of $Y$. Then $U_{\gamma}=Y\cap V_{\gamma}$ for each $\gamma\in\Gamma$, where $V_{\gamma}$ is $\tau_{X}$-open subset of $X$. Since $Y$ is $\tau_{X}$-closed in $X$, we see that
\[(X\setminus Y)\bigcup\left (\bigcup_{\gamma\in\Gamma}V_{\gamma}\right )\]
is an open cover of $X$. Therefore, there is a countable open subcover
\[(X\setminus Y)\bigcup\left (\bigcup_{n=1}^{\infty}V_{\gamma_{n}}\right )\]
of $X$. This also says that $\{Y\cap V_{\gamma_{n}}=U_{\gamma_{n}}\}_{n=1}^{\infty}$ is a countable open subcover of $Y$, which says that $Y$ is a Lindel\”{o}f space. This completes the proof. $\blacksquare$

The product space of a countable family of Lindelöf spaces is not necessarily a Lindelöf space. However, we have the following result.

Proposition. Suppose that the product space of a countable family of Lindelöf spaces is a Lindelöf space, and that each component space is a $T_{1}$-space. Then, each component space is also a Lindelöf space.

Proof. The result follows immediately from Proposition 68 and Proposition 69 in page topological spaces, and Proposition \ref{top49} in this page. $\blacksquare$

We have already known that the metric space satisfies the first axiom of countability. However the metric space fails to satisfy the second axiom of countability. The following result will show that, in metric space, the properties of being Lindelöf, separable and second countability are all equivalent.

Proposition. Let $(X,d)$ be a metric space. Then, the following statements are equivalent.

(a) $(X,d)$ is a Lindel\”{o}f space.

(b) $(X,d)$ is a separable space.

Proof. From Proposition 19 in page topological spaces and Proposition \ref{top50} in this page, we have already seen that the topological space satisfying the second axiom of countability is both a separable and Lindel\'{o}f space. Therefore it suffices to show that if $(X,d)$ is either a separable or Lindel\”{o}f space then it satisfies the second axiom of countability. We first show that statement (b) implies statement (c). Let $\{x_{n}\}_{n=1}^{\infty}$ be a countable dense subset of $X$. We shall show that
\[{\cal B}=\left\{B(x_{n};1/m):n,m\in\mathbb{N}\right\}\]
is a countable base for the metric topology on $X$. Let $O$ be any open subset of $X$, and let $x\in O$. There exists a positive number $r$ such that $B(x;r)\subseteq O$. Choose any integer $m>2/r$. Since $B(x;1/2m)$ is open and $\{x_{n}\}_{n=1}^{\infty}$ is dense. Proposition 20 in page topological spaces says that there exists $x_{n}$ in $B(x;1/2m)$, which also says $x\in B(x_{n};1/m)$. Since $m/2/r$, i.e., $1/m<r/2$, we have $B(x_{n};1/r)\subseteq B(x;r)$, which also says that $N(x_{n};1/m)\subseteq O$. Since $x$ can be any member of $O$, it follows that $O$ is the union of members of ${\cal B}$. This shows that ${\cal B}$ is a countable base for the metric topology on $X$; that is, $X$ satisfies the second axiom of countability.

Now, we wan to show that statement (a) implies statement (b). Suppose that $X$ is a Lindelöf space. Given $r>0$, using Zorn’s lemma, we can show that there exists a maximal subset of $X$ such that $d(a,b)\geq r$ for all $a,b\in E$. We want to claim that $E$ is countable. For each $a\in E$, we consider $B(a;r/2)$ and
\[V=X\setminus\bigcup_{a\in E}\mbox{cl}(B(a;r/4)).\]
Then $V$ is open and
\begin{equation}{\label{top51}}\tag{3}
\{V\}\cup\left\{B(a;r/2):a\in E\right\}
\end{equation}
is an open covering of $X$. Therefore, there is a countable open subcovering of $X$. If we remove $B(a;r/2)$ from the covering (\ref{top51}), then the remaining sets would fail to cover $X$, since none of them would contain $a$. Therefore the family $\{B(a;r/2):a\in E\}$ must be countable, which says that $E$ is indeed countable. According to the construction of $E$, we can construct $E_{n}$ for $r=1/n$ with $n=1,2,\cdots$, where $E_{n}$ is the maximal set such that $d(a,b)\geq r=\frac{1}{n}$ for any $a,b\in E_{n}$. Let $D=\bigcup_{n=1}^{\infty}$. Then $S$ is countable. Next we want to show that $D$ is dense in $X$. Given any $x\in X$ and $r>0$, we want to show that there exists $d\in D$ such that $d\in B(x;r)$. If $x\in D$ then it is done. For $x\not\in D$, i.e., $x\not\in E_{n}$ for all $n$, we take $n_{0}>1/r$. Suppose that every $y\in E_{n_{0}}$ satisfies $y\not\in B(x;r)$. This says that $d(x,y)\geq r>1/n_{0}$, which contradicts the maximality for $E_{n_{0}}$, since $x\not\in E_{n_{0}}$. Therefore, there exists $y_{0}\in E_{n_{0}}\subseteq D$ satisfying $y_{0}\in B(x;r)$. If $O$ is any nonempty open subset of $X$, then there exists $x\in O$ and $r>0$ satisfying $B(x;r)\subseteq O$, i.e., there exists $y_{0}\in D$ satisfying $y_{0}\in B(x;r)\subseteq O$. Proposition 20 in page topological spaces says that $D$ is dense in $X$, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top58}}\tag{4}\mbox{}\end{equation}
Proposition \ref{top58}. Suppose that $X$ is a $T_{3}$-space and Lindelöf space. Then, it is also a $T_{4}$-space.

Proof. Let $A$ and $B$ be disjoint closed subsets of $X$. If $x\in A$, then $X\setminus B$ is an open set containing $x$. By Proposition 40 in page topological spaces, there exists an open set $O_{x}$ containing $x$ satisfying $\mbox{cl}(O_{x})\subseteq X\setminus B$. Similarly, if $x\in B$ then there exists an open set $O_{x}$ containing $x$ such that $\mbox{cl}(O_{x})\subseteq X\setminus A$. If $x\not\in A\cup B$, then $X\setminus (A\cup B)$ is an open set containing $x$. Therefore, there also exists an open set $O_{x}$ containing $x$ satisfying $\mbox{cl}(O_{x})\subseteq X\setminus (A\cup B)$, i.e., $\mbox{cl}(O_{x})\cap (A\cup B)=\emptyset$. Therefore, the family of open sets $O_{x}$ for each $x$ described above is an open covering of $X$. Since $X$ is a Lindel\”{o}f space, there is a countable open subcovering of $X$. Let $\{O_{n}^{(A)}\}_{n=1}^{\infty}$ be the open sets which meet $A$, and let $\{O_{n}^{(B)}\}_{n=1}^{\infty}$ be the open sets which meet $B$. Then $\mbox{cl}(O_{n}^{(A)})\cap B=\emptyset$ and $\mbox{cl}(O_{n}^{(B)})\cap A=\emptyset$ for all $n$. We also have $A\subset\bigcup_{n\in\mathbb{N}}O_{n}^{(A)}$ and $B\subset\bigcup_{n\in\mathbb{N}}O_{n}^{(B)}$. Define $W_{1}=O_{1}^{(A)}$ and set $Y_{1}=O_{1}^{(B)}\setminus\mbox{cl}(W_{1})$. Let $W_{2}=O_{2}^{(A)}\setminus\mbox{cl}(Y_{1})$ and $Y_{2}=O_{2}^{(B)}\setminus (\mbox{cl}(W_{1})\cup\mbox{cl}(W_{2}))$. Inductively, if $W_{n}$ and $Y_{n}$ are defined, then $W_{n+1}$ and $Y_{n+1}$ are defined in the following way:
\[W_{n+1}=O_{n+1}^{(A)}\setminus\bigcup_{i=1}^{n}\mbox{cl}(Y_{i})\mbox{ and }
Y_{n+1}=O_{n+1}^{(B)}\setminus\bigcup_{i=1}^{n+1}\mbox{cl}(W_{i}).\]
Since
\[W_{n}=O_{n}^{(A)}\cap\left (X\setminus\bigcup_{i=1}^{n-1}\mbox{cl}(Y_{i})\right )
=O_{n}^{(A)}\cap\left (X\setminus\mbox{cl}\left (\bigcup_{i=1}^{n-1}Y_{i}\right )\right ),\]
it says that $W_{n}$ is open. We can similarly show that $Y_{n}$ is open. Let
\[H=\bigcup_{n\in\mathbb{N}}W_{n}\mbox{ and }K=\bigcup_{n\in\mathbb{N}}Y_{n}.\]
Then $H$ and $K$ are also open. Suppose that $x\in A$. Then $x\in O_{n}^{(A)}$ for some $n$. Since $\mbox{cl}(Y_{i})\subset\mbox{cl}(O_{i}^{(B)})$ and $\mbox{cl}(O_{i}^{(B)})\cap A=\emptyset$, it follows $a\not\in\mbox{cl}(Y_{i})$ for each $i$. Since
\[W_{n}=O_{n}^{(A)}\setminus\bigcup_{i=1}^{n-1}\mbox{cl}(Y_{i}),\]
we obtain $a\in W_{n}$, which also says that $A\subseteq H$. We can similarly show that $B\subseteq K$. In order to show that $X$ is a $T_{4}$-space, we remain to show $H\cap K=\emptyset$. Suppose that $x\in H\cap K$. Then $x\in W_{n}\cap Y_{m}$ for some $n$ and $m$.

Assume that $m\geq n$. Then, we have
\[x\in Y_{m}=O_{m}^{(B)}\setminus\left (\mbox{cl}(W_{1})\cup\cdots\cup\mbox{cl}(W_{n})\cup\cdots\cup\mbox{cl}(W_{m})\right ),\]
which says that $x\not\in\mbox{cl}(W_{n})$. A contradiction occurs.
Assume that $m<n$. Then, we have
\[x\in W_{n}=O_{n}^{(A)}\setminus\left (\mbox{cl}(Y_{1})\cup\cdots\cup\mbox{cl}(Y_{m})\cup\cdots\cup\mbox{cl}(Y_{n-1})\right ),\]
which says that $x\not\in\mbox{cl}(Y_{m})$. A contradiction also occurs.

This completes the proof. $\blacksquare$

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

Compactness.

Definition. A topological space $X$ is said to be compact when every open covering of $X$ has a finite subcovering. A subset $K$ of a topological space is called compact when it is compact as a subspace of $X$. $\sharp$

It is clear to see that $\{O_{\gamma}\}_{\gamma\in\Gamma}$ is a $\tau_{X}$-open covering of the set $K$ if and only if $\{O_{\gamma}\cap K\}_{\gamma\in\Gamma}$ is a $\tau_{K}$-open covering of the subspace $K$. Therefore the subspace $K$ is compact if and only if there is a finite $\tau_{K}$-open subcovering $\{O_{\gamma_{i}}\cap K\}_{i=1}^{n}$ of $K$, i.e., $\{O_{\gamma_{i}}\}_{i=1}^{n}$ is a finite $\tau_{X}$-open subcovering of $K$. In other words, a subset $K$ of $X$ is compact if and only if every $\tau_{X}$-open covering of $K$ has a finite $\tau_{X}$-open subcovering. We also have the following observations.

The Heine-Borel Theorem states that every closed and bounded subset of $\mathbb{R}^{n}$ is compact
${\cal O}$ is an open covering of the space $X$ if and only if the collection ${\cal C}$ of complements of sets in ${\cal O}$ is a collection of closed sets whose intersection is empty. In other words, a space $X$ is compact if and only if every collection of closed sets with an empty intersection has a finite subcollection whose intersection is empty.

\begin{equation}{\label{top73}}\tag{5}\mbox{}\end{equation}
Proposition \ref{top73}. Let $(Y,\tau_{Y})$ be a subspace of the topological space $(X,\tau_{X})$. Then, we have the following properties.

(i) Suppose that $K$ is $\tau_{X}$-compact and $K\subseteq Y$. Then $K$ is $\tau_{Y}$-compact.

(ii) Suppose that $K$ is $\tau_{Y}$-compact. Then $K$ is $\tau_{X}$-compact.

Proof. To prove part (i), let $\{O_{\gamma}^{*}\}_{\gamma\in\Gamma}$ be a $\tau_{Y}$-open covering of $K$, i.e., $K\subseteq\bigcup_{\gamma\in\Gamma}O_{\gamma}^{*}$. Then, we have $O_{\gamma}^{*}=Y\cap O_{\gamma}$ for some $\tau_{X}$-open set $O_{\gamma}$. Therefore, we obtain
\[K\subseteq\bigcup_{\gamma\in\Gamma}O_{\gamma}^{*}=\bigcup_{\gamma\in\Gamma}
\left (Y\cap O_{\gamma}\right )=Y\cap\left (\bigcup_{\gamma\in\Gamma}O_{\gamma}\right )
\subseteq\bigcup_{\gamma\in\Gamma}O_{\gamma},\]
which implies $K\subseteq\bigcup_{i=1}^{n}O_{\gamma_{i}}$ by the $\tau_{X}$-compactness. Now, we also obtain
\[K=K\cap Y\subseteq Y\cap\left (\bigcup_{i=1}^{n}O_{\gamma_{i}}\right )
=\bigcup_{i=1}^{n}\left (Y\cap O_{\gamma_{i}}\right )=\bigcup_{i=1}^{n}O_{\gamma_{i}}^{*},\]
which shows that $K$ is $\tau_{Y}$-compact.

To prove part (ii), let $\{O_{\gamma}\}_{\gamma\in\Gamma}$ be a $\tau_{X}$-open covering of $K$, i.e., $K\subseteq\bigcup_{\gamma\in\Gamma}O_{\gamma}$. Since $K\subseteq Y$, we have
\[K=K\cap Y\subseteq Y\cap\left (\bigcup_{\gamma\in\Gamma}O_{\gamma}\right )
=\bigcup_{\gamma\in\Gamma}\left (Y\cap O_{\gamma}\right )=\bigcup_{\gamma\in\Gamma}O_{\gamma}^{*},\]
where each $O_{\gamma}^{*}$ is $\tau_{Y}$-open. By the $\tau_{Y}$-compactness, we have
\[K\subseteq\bigcup_{i=1}^{n}O_{\gamma_{i}}^{*}=\bigcup_{i=1}^{n}\left (Y\cap O_{\gamma_{i}}\right )
=Y\cap\left (\bigcup_{i=1}^{n}O_{\gamma_{i}}\right )\subseteq\bigcup_{i=1}^{n}O_{\gamma_{i}},\]
which shows that $K$ is $\tau_{X}$-compact. This completes the proof. $\blacksquare$

Definition. A family ${\cal A}$ of sets has the finite intersection property when the intersection of the members of each finite subfamily of ${\cal A}$ is nonempty. $\sharp$

\begin{equation}{\label{top52}}\tag{6}\mbox{}\end{equation}
Proposition \ref{top52}. A topological space is compact if and only if each family of closed sets having the finite intersection property has a nonempty intersection.

Proof. Suppose that $X$ is compact. Let $\{C_{\gamma}\}_{\gamma\in\Gamma}$ be any family of closed subsets of $X$ satisfying $\bigcap_{\gamma\in\Gamma}C_{\gamma}=\emptyset$. We want to lead to a contradiction. Let $O_{\gamma}=X\setminus C_{\gamma}$ for $\gamma\in\Gamma$. Then, we have
\[X=X\setminus\bigcap_{\gamma\in\Gamma}C_{\gamma}=\bigcup_{\gamma\in\Gamma}
\left (X\setminus C_{\gamma}\right )=\bigcup_{\gamma\in\Gamma}O_{\gamma}.\]
Therefore $\{O_{\gamma}\}_{\gamma\in\Gamma}$ is an open covering of $X$. The compactness of $X$ says that there exists finitely many $O_{\gamma_{1}},\cdots ,O_{\gamma_{n}}$ satisfying $X=\bigcup_{i=1}^{n}O_{\gamma_{i}}$. Therefore, we obtain $\bigcap_{i=1}^{n}C_{\gamma_{i}}=\emptyset$, which contradicts the finite intersection property.

To prove the converse, given any family $\{C_{\gamma}\}_{\gamma\in\Gamma}$ of closed subsets of $X$, we assume that if the intersection of all members in $\{C_{\gamma}\}_{\gamma\in\Gamma}$ is empty then there exist finitely many $C_{\gamma_{1}},\cdots ,C_{\gamma_{n}}$ satisfying $\bigcap_{i=1}^{n}C_{\gamma_{i}}=\emptyset$. Let $\{O_{\gamma}\}_{\gamma\in\Gamma}$ be any open covering of $X$. Let $C_{\gamma}=X\setminus O_{\gamma}$ for $\gamma\in\Gamma$. Then $\{C_{\gamma}\}_{\gamma\in\Gamma}$ is a family of closed subsets of $X$ such that the intersection of all members in $\{C_{\gamma}\}_{\gamma\in\Gamma}$ is empty. Therefore, there exist finitely many $C_{\gamma_{1}},\cdots ,C_{\gamma_{n}}$ satisfying $\bigcap_{i=1}^{n}C_{\gamma_{i}}=\emptyset$, which also says $X=\bigcup_{i=1}^{n}O_{\gamma_{i}}$. This shows that $X$ is compact, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top57}}\tag{7}\mbox{}\end{equation}
Proposition \ref{top57}. Let $(K,\tau_{K})$ be a compact subspace of a topological space $(X,\tau )$, and let $F$ be a subset of $K$. Suppose that $F$ is $\tau$-closed or $\tau_{K}$-closed. Then, it is both $\tau_{K}$-compact and $\tau$-compact.

Proof. Let ${\cal O}=\{O_{\gamma}\}_{\gamma\in\Gamma}$ be a $\tau_{K}$-open covering of $F$. Suppose that $F$ is $\tau$-closed and $F\subseteq K$. Proposition 15 in page topological spaces says that $F$ is also $\tau_{K}$-closed. It follows that ${\cal O}\cup\{K\setminus F\}$ is a $\tau_{K}$-open
covering of $K$. Therefore, we must have a finite subcovering $\{K\setminus F,O_{\gamma_{1}},\cdots ,O_{\gamma_{n}}\}$ of $K$, which says that $\{O_{\gamma_{1}},\cdots ,O_{\gamma_{n}}\}$ covers $F$. This shows that $F$ is $\tau_{K}$-compact. Proposition \ref{top73} says that $F$ is also $\tau$-compact. This completes the proof. $\blacksquare$

\begin{equation}{\label{top59}}\tag{8}\mbox{}\end{equation}
Proposition \ref{top59}. A compact subset of a Hausdorff space is closed.

Proof. Let $X$ be a Hausdorff space, and let $K$ be a compact subset of $X$. We shall show that $X\setminus K$ is open. Given any $x\in X\setminus K$, we shall claim that there exists a neighborhood $N$ of $x$ satisfying $N\subseteq X\setminus K$. Since $X$ is Hausdorff, for each $y\in K$, there exist disjoint open sets $O_{y}^{(1)}$ and $O_{y}^{(2)}$ containing $x$ and $y$, respectively. The collection of sets $\{O_{y}^{(2)}:y\in K\}$ form an open covering of $K$. Therefore, there is a finite subcollection $\{O_{y_{1}}^{(2)},\cdots ,O_{y_{n}}^{(2)}\}$ which covers $K$. Let
\[O^{(1)}=\bigcap_{i=1}^{n}O_{y_{i}}^{(1)}\mbox{ and }O^{(2)}=\bigcup_{i=1}^{n}O_{y_{i}}^{(2)}.\]
Then $x\in O^{(1)}$ and $K\subseteq O^{(2)}$ satisfying $O^{(1)}\cap O^{(2)}=\emptyset$, which also says that $O^{(1)}\subseteq X\setminus K$. Since $O^{(1)}$ is open, i.e., a neighborhood of $x$, it follows that $X\setminus K$ is open, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top60}}\tag{9}\mbox{}\end{equation}
Corollary \ref{top60}. Let $(X,\tau )$ be a compact Hausdorff space. Then a subset of $X$ is compact if and only if
it is closed.

Proof. This is obvious from Propositions \ref{top57} and \ref{top59}. $\blacksquare$

Corollary. Every compact set of real numbers is closed and bounded.

Proof. Since $\mathbb{R}$ is a Hausdorff space, it follows that a compact subset $K$ of $\mathbb{R}$ must be closed. Since the collection of open intervals $\{(-r,r):r\in\mathbb{R}\}$ form an open covering of $K$. Therefore, a finite number of them must cover $K$, which says that $K$ must be bounded. This completes the proof. $\blacksquare$

\begin{equation}{\label{top69}}\tag{10}\mbox{}\end{equation}
Proposition \ref{top69}. A compact Hausdorff space is both a $T_{3}$-space and $T_{4}$-space.

Proof. Let $F$ be a closed subset of a compact Hausdorff space $X$. Corollary \ref{top60} says that $F$ is compact. Given any $x\in X\setminus F$, from the proof of Proposition \ref{top59}, there exists disjoint open sets $O^{(1)}$ and $O^{(2)}$ satisfying $x\in O^{(1)}$ and $F\subseteq O^{(2)}$, which shows that $X$ is a $T_{3}$-space. Finally, since a compact space is also a Lindelöf space. Proposition \ref{top58} says that a compact Hausdorff space is also a $T_{4}$-space. This completes the proof. $\blacksquare$

\begin{equation}{\label{top82}}\tag{11}\mbox{}\end{equation}
Proposition \ref{top82}. Let $(Y,\tau_{Y})$ be a closed subspace of a topological space $(X,\tau_{X})$. Suppose that $K$ is a $\tau_{X}$-compact. Then $K\cap Y$ is both $\tau_{X}$-compact and $\tau_{Y}$-compact.

Proof. Considering the subspace $(K,\tau_{K})$, since $Y$ is $\tau_{X}$-closed, Proposition 15 in page topological spaces says that $Y\cap K$ is $\tau_{K}$-closed. Proposition \ref{top57} also says that $Y\cap K$ is $\tau_{K}$-compact. Using part (ii) of Proposition \ref{top73}, it follows that $Y\cap K$ is $\tau$-compact. Also, using part (i) of Proposition \ref{top73}, we conclude that $Y\cap K$ is $\tau_{Y}$-compact. This completes the proof. $\blacksquare$

\begin{equation}{\label{top61}}\tag{12}\mbox{}\end{equation}
Proposition \ref{top61}. Let $f$ be a continuous function from the topological space $(X,\tau_{X})$ into the topological space $(Y,\tau_{Y})$. Suppose that $K$ is a $\tau_{X}$-compact subset of $X$. Then, the image $f(K)$ is a $\tau_{Y}$-compact subset of $Y$.

Proof. Let ${\cal O}=\{O_{\gamma}\}_{\gamma\in\Gamma}$ be a $\tau_{Y}$-open covering for $f(K)$. Then, the collection of sets
\[f^{-1}({\cal O})=\left\{f^{-1}(O_{\gamma})\right\}_{\gamma\in\Gamma}\]
is also a $\tau_{X}$-open covering of $K$ using the continuity. Since $K$ is $\tau_{X}$-compact, there are finite numbers $f^{-1}(O_{1}),\cdots ,f^{-1}(O_{n})$ which cover $K$. This also says that the sets $O_{1},\cdots ,O_{n}$ cover $f(K)$, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top255}}\tag{13}\mbox{}\end{equation}
Proposition \ref{top255}. Suppose that $X$ is a Lindel\”{o}f space and every sequence in $X$ has a cluster point. Then $X$ is compact.

Proof. It must be shown that each open cover of $X$ has a finite subcover. Because of the hypothesis, it may be assumed that the open cover consists of sets
$A_{0},A_{1},\cdots ,A_{n},\cdots$. Proceeding inductively, let $B_{0}=A_{0}$ and for each $p$, let $B_{p}$ be the first of the sequence of $A$’s which
is not covered by $B_{0}\cup B_{1}\cup\cdots\cup B_{p-1}$. If this choice is impossible at any stage, then the sets already selected are the required finite subcover. Otherwise it is possible to select a point $b_{p}\in B_{p}$ for each $p$ such that $b_{p}\not\in B_{i}$ for $i<p$. Let $x$ be a cluster point of this sequence. Then $x\in B_{p}$ for some $p$, and since $x$ is a cluster point, $b_{q}\in B_{p}$ for some $q>p$, which is a contradiction. $\blacksquare$

\begin{equation}{\label{top54}}\tag{14}\mbox{}\end{equation}
Proposition \ref{top54}. A topological space $X$ is compact if and only if every net in $X$ has a cluster point

Proof. Suppose that $X$ is compact. Let $\{x_{\alpha}\}_{\alpha\in\Gamma}$ be any net in $X$. Define $B_{\beta}=\{x_{\alpha}:\beta\prec\alpha\}$. Then, the intersection of any finite family of the sets $\mbox{cl}(B_{\beta})$ is nonempty. Proposition \ref{top52} says $\bigcap_{\beta\in\Gamma}\mbox{cl}(B_{\beta})\neq\emptyset$. Let $x$ be an element in $\bigcap_{\beta\in\Gamma}\mbox{cl}(B_{\beta})$. We want to claim that $x$ is a cluster point of the net $\{x_{\alpha}\}_{\alpha\in\Gamma}$. Since $x\in\mbox{cl}(B_{\beta})$ for any $\beta\in\Gamma$, any neighborhood $N_{x}$ of $x$ meets $B_{\beta}$. Given any $\beta_{1},\beta_{2}\in\Gamma$, since $\Gamma$ is a directed set, there exists $\beta_{3}\in\Gamma$ such that $\beta_{1}\prec\beta_{3}$ and $\beta_{2}\prec\beta_{3}$. Since $N_{x}\cap B_{\beta_{3}}\neq\emptyset$, let $x_{\beta_{0}}\in N_{x}\cap B_{\beta_{3}}$. Then, by the definition of $B_{\beta_{3}}$, it follows that $\beta_{0}\in\Gamma$ satisfying $\beta_{3}\prec\beta_{0}$. In other words, given any $N_{x}$ and any $\beta_{1}\in\Gamma$, there exists $\beta_{0}\succ\beta_{1}$ satisfying $x_{\beta_{0}}\in N_{x}$. This shows that $x$ is indeed a cluster point of the net $\{x_{\alpha}\}_{\alpha\in\Gamma}$.

For the converse, let $\{C_{\delta}\}_{\delta\in\Theta}$ be the family of closed subset of $X$ having the finite intersection property. Let $\Gamma$ be the family of all finite intersection of members in $\{C_{\delta}\}_{\delta\in\Theta}$. Then $\Gamma$ is a directed set in the sense of $A\prec B$ when $B\subseteq A$. Since each member in $\Gamma$ is nonempty, we can form a set
\[\left\{x_{A}:A\in\Gamma\mbox{ and }x_{A}\in A\right\}.\]
Then $\{x_{A}\}_{A\in\Gamma}$ is a net in $X$. The assumption says that the net $\{x_{A}\}_{A\in\Gamma}$ has a cluster point $x$. Given any $C_{\delta}$, if $A\in\Gamma$ and $A\succ C_{\delta}$, i.e., $A\subseteq C_{\delta}$, then $x_{A}\in A\subseteq C_{\delta}$. Proposition \ref{top53} says that there exists a subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of $\{x_{A}\}_{A\in\Gamma}$ such that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ converges to $x$. Now, according to the definition of subnet, there exists $\beta\in\Gamma$ such that if
$\gamma\succ\beta$ then $N_{\gamma}\succ A$. The transitivity says $N_{\gamma}\succ C_{\delta}$. Therefore, we have $x_{N_{\gamma}}\in C_{\delta}$. Since $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ converges to $x$, given any neighborhood $N_{x}$ of $x$, there exists $\beta_{0}\in\Gamma$ such that $\gamma\succ\beta_{0}$ implies $x_{N_{\gamma}}\in N_{x}$. Therefore, we must have $N_{x}\cap C_{\delta}\neq\emptyset$, which says $x\in\mbox{cl}(C_{\delta})$. Since each $C_{\delta}$ is closed, we obtain $x\in C_{\delta}$ for all $\delta\in\Theta$; that is, the intersection of all members $C_{\delta}$ is nonempty. Proposition \ref{top52} says that $X$ is compact. This completes the proof. $\blacksquare$

Proof. (Second Proof). Let $\{S_{n}:n\in D\}$ be a net in the compact topological space $X$ and for each $n\in D$, let $A_{n}$ be the set of all points $S_{m}$ for $m\geq n$. Then, the family of all sets $A_{n}$ has the finite intersection property since $D$ is directed by $\geq$, and consequently the family of all closures $\mbox{cl}(A_{n})$ also has the finite intersection property. Since $X$ is compact, there is a point $s$ which belongs to each $\mbox{cl}(A_{n})$ by Proposition \ref{top52}, and according to Proposition 53 in page topological spaces such a point $s$ is a cluster point of the net $\{S_{n}:n\in D\}$. To prove the converse, let $X$ be a topological space in which every net has a cluster point and let ${\cal A}$ be a family of closed subsets of $S$ such that ${\cal A}$ has the finite intersection property. Define ${\cal B}$ to be the family of all finite intersections of members of ${\cal A}$; then ${\cal B}$ has the finite intersection property and since ${\cal A}\subseteq {\cal B}$, it is sufficient to show $\bigcap_{B\in {\cal B}}B$ nonempty. The intersection of two members of ${\cal B}$ is a member of ${\cal B}$ and therefore ${\cal B}$ is directed by $\subseteq$. If we choose a member $S_{B}$ from each $B\in {\cal B}$, then $\{S_{B}:B\in {\cal B}\}$ is a net in $X$ and consequently has a cluster point $s$. If $B$ and $C$ are members of ${\cal B}$ such that $C\subseteq B$, then $S_{c}\in C\subseteq B$; therefore the net $\{S_{B}:B\in {\cal B}\}$ is eventually in the closed set $B$ and hence the cluster point $s$ belongs to $B$. Therefore $s$ belongs to each member of ${\cal B}$ and the intersection of the members of ${\cal B}$ is nonempty. Finally, the second statement follows from Proposition 51 in page topological spaces. $\blacksquare$

\begin{equation}{\label{top63}}\tag{15}\mbox{}\end{equation}
Proposition \ref{top63}. A topological space $X$ is compact if and only if every ultranet in $X$ is convergent.

Proof. Suppose that $X$ is compact. Proposition \ref{top54} says that every ultranet in $X$ has a cluster point. Using Proposition 64 in page topological spaces, we see that every ultranet in $X$ is convergent. For the converse, given any net $\{x_{\alpha}\}_{\alpha\in\Gamma}$ in $X$, Proposition 63 in page topological spaces says that there exists a subnet $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ of $\{x_{\alpha}\}_{\alpha\in\Gamma}$ such that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is an ultranet. The assumption says that $\{x_{N_{\beta}}\}_{\beta\in\Gamma}$ is convergent. Proposition 51 in page topological spaces says that $x$ is a cluster point of $\{x_{\alpha}\}_{\alpha\in\Gamma}$. Using Proposition \ref{top54}, we complete the proof. $\blacksquare$

Proposition. Let $f$ be a continuous and one-to-one function from a compact space $(X,\tau_{X})$ onto a Hausdorff space $(Y,\tau_{Y})$. Then $f^{-1}$ is continuous; that is to say, the function $f$ is a homeomorphism.

Proof. Let $F$ be any $\tau_{X}$-closed subset of $X$. Proposition \ref{top57} says that $F$ is also a compact subset of $X$. Using Proposition \ref{top61}, we also see that $f(F)$ is a compact subset of $Y$. From Proposition \ref{top59}, it follows that $f(F)$ is $\tau_{Y}$-closed subset of $Y$. Since $f(F)=(f^{-1})^{-1}(F)$, Proposition 24 in page topological spaces concludes that $f^{-1}$ is continuous, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top66}}\tag{16}\mbox{}\end{equation}
Proposition \ref{top66}. $(X,\tau)$ is a compact space if and only if there exists a subbase ${\cal S}$ such that if ${\cal O}$ is an open cover of $X$ consisting of elements of ${\cal S}$ then ${\cal O}$ contains a finite subcover of $X$.

Proof. If $(X,\tau )$ is compact, then we can take ${\cal S}=\tau$ which satisfies the required property. For the converse, suppose that $X$ has a subbase ${\cal S}$ having the required property. Let ${\cal A}\subseteq\tau$ and ${\cal A}$ does not contain a finite subcover of $X$. We shall prove that ${\cal A}$ cannot be a cover of $X$, which indirectly shows that any open cover of $X$ contains a finite subcover. Let $\mathfrak{H}$ be the collection of all ${\cal C}\subseteq\tau$ such that ${\cal A}\subseteq {\cal C}$ and ${\cal C}$ does not contain a finite subcover of $X$. Then $\mathfrak{H}\neq\emptyset$ since ${\cal A}\in\mathfrak{H}$. We also see that the inclusion “$\subset$” is a partial ordering of $\mathfrak{H}$. Let $\mathfrak{K}$ be any chain in the partially ordered set $(\mathfrak{H},\subseteq )$. Then, the union of members of $\mathfrak{K}$ is a member of $\mathfrak{H}$ and is an upper bound for $\mathfrak{K}$. Zorn’s lemma says that $\mathfrak{H}$ contains a maximal element ${\cal M}$, which also says
that ${\cal A}\subseteq {\cal M}$ and ${\cal M}$ does not contain a finite subcover of $X$. Since ${\cal A}\subseteq {\cal M}$, if we can show that ${\cal M}$ is not a cover of $X$ then ${\cal A}$ cannot be a cover of $X$.

We assume that ${\cal M}$ is a cover of $X$ to get a contradiction. For each $x\in X$, there exists $M$ satisfying $x\in M\in {\cal M}\subseteq\tau$. Since $M\in\tau$ and ${\cal S}$ is a subbase, there exist finitely many members $S_{1},\cdots ,S_{n}$ of ${\cal S}$ satisfying
\[x\in S_{1}\cap\cdots\cap S_{n}\subseteq M.\]
Suppose that $S_{i}\not\in {\cal M}$ for each $i=1,\cdots ,n$. Since ${\cal M}$ is a maximal element of $\mathfrak{H}$, if ${\cal M}\cup\{S_{i}\}$ does not contain a finite subcover of $X$, then it contradicts the maximality. Therefore, we must have that ${\cal M}\cup\{S_{i}\}$ contains a finite subcover of $X$ for each $i=1,\cdots ,n$. This also says that there exists a finite subcollection ${\cal M}_{i}$ of ${\cal M}$ such that ${\cal M}_{i}\cup\{S_{i}\}$
is a finite cover of $X$. Since $S_{1}\cap\cdots\cap S_{n}\subseteq M$, it follows
\[\{M\}\cup {\cal M}_{1}\cup\cdots\cup {\cal M}_{n}\subseteq {\cal M}\]
forms a finite subcover of $X$. This is a contradiction, since ${\cal M}$ does not contain any finite subcover of $X$. Therefore, there exists $S_{i_{0}}\in {\cal M}$ satisfying $x\in S_{i_{0}}$. The above argument shows that, given any $x\in X$, there exists $S\in {\cal S}$ satisfying $x\in S\in {\cal M}$, i.e., $x\in S\in {\cal S}\cap {\cal M}$. This says that ${\cal S}\cap {\cal M}$ is an open cover of $X$. Since ${\cal S}\cap {\cal M}\subseteq {\cal S}$, according to the assumption, we must have that ${\cal S}\cap {\cal M}$ contains a finite subcover of $X$. Since ${\cal S}\cap {\cal M}\subseteq {\cal M}$, it also says that ${\cal M}$ contains a finite subcover of $X$. This contradiction says that ${\cal M}$ cannot be a cover of $X$; that is, ${\cal A}$ cannot be a cover of $X$. This completes the proof. $\blacksquare$

Definition. A point $x$ is an $\omega$-accumulation point of a set $A$ when each neighborhood of $x$ contains infinitely many points of $A$. $\sharp$

Each $\omega$-accumulation point of a set is also an accumulation point. If the space is $T_{1}$, then the converse also holds.

\begin{equation}{\label{top254}}\tag{17}\mbox{}\end{equation}
Proposition \ref{top254}. Every sequence in a topological space has a cluster point if and only if every infinite set has an $\omega$-accumulation point.

Proof. Suppose that every sequence has a cluster point and that $A$ is an infinite subset. Then, there is a sequence of distinct points in $A$, and each cluster point of such a sequence is clearly an $\omega$-accumulation point of $A$. Conversely, if every infinite subset of a topological space has an accumulation point and $\{S_{n}\}$ is a sequence in the space, then one of two situations must occur. Either the range of the sequence is infinite, in which case each $\omega$-accumulation point of this infinite set is a cluster point of the sequence, or else the range of the sequence is finite. In the latter case, for some point $x$ of the space, $S_{n}=x$ for infinitely many nonnegative integers $n$, and $x$ is a cluster point of the sequence. $\blacksquare$

Theorem. Let $X$ be a topological space. We consider the following four statements

(a) Every infinite subset of $X$ has an $\omega$-accumulation point.

(b) Every sequence in $X$ has a cluster point.

(d) The space $X$ is compact.

Then, we have the following properties.

(i) (a) is equivalent to (b), and (d) implies (a).

(ii) If $X$ satisfies the first axiom of countability, then (a), (b) and (c) are equivalent.

(iii) If $X$ satisfies the second axiom of countability, then all four statements are equivalent.

(iv) If $X$ is a pseudo-metric space, then each of the four statements implies that $X$ satisfies the second axiom of countability and all four statements are equivalent.

Proof. Proposition \ref{top254} states that (a) is equivalent to (b) and since a sequence is a net, Proposition \ref{top54} shows that (d) always implies (b). If $X$ satisfies the first axiom of countability then (b) and (c) are equivalent by Proposition 54 in page topological spaces. If $X$ satisfies the second axiom of countability, then every open cover has a countable subcover, Proposition \ref{top255} applies, and hence all four statements are equivalent. If $X$ is pseudo-metric, then $X$ satisfies the first axiom of countability, the first three conditions are equivalent, each is implied by compactness, and the theorem will be proved if it is shown that a pseudo-metric space such that each infinite subset has an accumulation point is separable and hence satisfies the second axiom of countability. Suppose that $X$ is such a pseudo-metric space. For $r$ positive consider the family of all sets $A$ such that the distance between any two distinct points of $A$ is at least $r$. It is easily seen that this family has a maximal member $A_{r}$ by $0.25$. The set $A_{r}$ must be finite, for the $r/2$-sphere about each point of $X$ contains at most one member of $A_{r}$ and therefore $A_{r}$ has no accumulation point. Moreover, the $r$-sphere about each point $x$ of $X$ must intersect $A_{r}$ because $A_{r}$ is maximal and otherwise $x$ could be adjoined to $A_{r}$. Finally the union $A$ of sets $A_{r}$, for $r$ the reciprocal of a positive integer, is surely countable and $A$ is clearly dense in $X$. $\blacksquare$

If ${\cal B}$ is a base for the topology of a compact space $X$ and ${\cal A}$ is a cover of $X$ by members of ${\cal B}$, then there is a finite subcover of ${\cal A}$. Conversely, suppose that ${\cal B}$ is a base for the topology and that every cover by members of ${\cal B}$ has a finite subcover. If ${\cal C}$ is an arbitrary open cover of $X$, we define ${\cal A}$ to be the family of all members of ${\cal B}$ which are subsets of some members of ${\cal C}$. Because ${\cal B}$ is a base, the family ${\cal A}$ is a cover of $X$ (since every member of ${\cal C}$ is the union of members of ${\cal B}$, i.e., every member of ${\cal B}$ is a subset of some members of ${\cal C}$), and consequently there is a finite subcover ${\cal A}’$ of ${\cal A}$. For each member of ${\cal A}’$ we may select a member of ${\cal C}$ which contains it, and the result is a finite subcover of ${\cal C}$. This shows that $X$ is compact. This is useful but not a very profound result. The corresponding theorem on subbase is both profound and useful.

Definition. A family ${\cal A}$ of sets is of finite character when each finite subset of a member of ${\cal A}$ is a member of ${\cal A}$, and each set $A$, every finite subset of which belongs to ${\cal A}$, itself belongs to ${\cal A}$.

\begin{equation}{\label{top256}}\tag{18}\mbox{}\end{equation}
Lemma \ref{top256}. (Tukey Lemma). There is a maximal member of each nonempty famil of finite character. $\sharp$

\begin{equation}{\label{top257}}\tag{19}\mbox{}\end{equation}
Theorem \ref{top257}. (Alexander). If ${\cal S}$ is a subbase for the topology of a space $X$ such that every cover of $X$ by members of ${\cal S}$ has a finite subcover, then $X$ is compact.

Proof. For brevity let us agree that a family of subsets of $X$ is inadequate if and only if it fails to cover $X$, and is finitely inadequate if and only if no finite subfamily covers $X$. Then, the definition of compactness of $X$ can be stated: each finitely inadequate family of open sets is inadequate. Observe that the class of finitely inadequate families of open sets is of finite character and therefore each finitely inadequate family is contained in a maximal family by Tukey’s Lemma \ref{top256}. Such a maximal finitely inadequate family ${\cal A}$ has a special property which is established as follows. If $C\not\in {\cal A}$ and $C$ is open, then by maximality there is a finite subfamily $A_{1},\cdots ,A_{m}$ of ${\cal A}$ satisfying $C\cup A_{1}\cup\cdots\cup A_{m}=X$. Hence no open set containing $C$ belongs to ${\cal A}$. If $D$ is another open set and $D\not\in {\cal A}$, then there is $B_{1},\cdots ,B_{n}$ in ${\cal A}$ satisfying \[D\cup B_{1}\cup\cdots\cup B_{n}=X\] and \[(C\cap D)\cup A_{1}\cup\cdots\cup A_{m}\cup B_{1}\cup\cdots\cup B_{n}=X.\] It follows $C\cap D\not\in {\cal A}$. Consequently, if no member of a finite family of open sets belongs to ${\cal A}$, then no open set containing the intersection belongs to ${\cal A}$; restated, if a member of ${\cal A}$ contains a finite intersection $C_{1}\cap\cdots\cap C_{p}$ of open sets, then some $C_{i}\in {\cal A}$. The proof is now straightforward. Suppose that ${\cal S}$ is a subbase such that each open cover by subbase elements has a finite subcover (that is, each finitely inadequate subfamily is inadequate) and suppose that ${\cal B}$ is a finitely inadequate family of open subsets of $X$. Then, there is a maximal family ${\cal A}$ of this sort containing ${\cal B}$ and it is sufficient to show that ${\cal A}$ is inadequate. The family ${\cal S}\cap {\cal A}$ of all members of ${\cal A}$ which belongs to ${\cal S}$ is finitely inadequate and hence ${\cal S}\cap {\cal A}$ does not cover $X$. Consequently, the result will be proved if it is shown that each point in $\bigcup_{A\in {\cal A}}A$ belongs to $\bigcup_{A\in ({\cal S}\cap {\cal A})}A$. Because ${\cal B}$ is a subbase, each point $x$ of a member $A$ of ${\cal A}$ belongs to some finite intersection of members of ${\cal S}$ which is contained in $A$. The paragraph above shows that some of this finite family belongs to ${\cal A}$, hence

\[\bigcup_{A\in {\cal A}}A=\bigcup_{A\in ({\cal S}\cap {\cal A})}A,\]

and this completes the proof. $\blacksquare$

Let $X$ be a compact Hausdorff space. We denote by ${\cal C}(X)$ the set of all continuous real-valued functions defined on $X$. Then, we have the following observations.

The set ${\cal C}(X)$ is a vector space, since any constant multiple of a continuous real-valued function is continuous, and the sum of two continuous functions is continuous.
The space ${\cal C}(X)$ becomes a normed vector space when we define
\[\parallel f\parallel =\max_{x\in X}|f(x)|.\]
The space ${\cal C}(X)$ is also a metric space when we define $d(f,g)=\parallel f-g\parallel$. Moreover ${\cal C}(X)$ is a complete metric space.
Since $X$ is a normal space, given any two distinct points $x$ and $y$ in $X$, the Urysohn’s Lemma 42 in page topological spaces says that there exists $f\in {\cal C}(X)$ satisfying $f(x)\neq f(y)$.

Given any $f,g\in {\cal C}(X)$, we define the functions $f\wedge g$ and $f\vee g$ by
\[(f\wedge g)(x)=\min\{f(x),g(x)\}\]

and

\[(f\vee g)(x)=\max\{f(x),g(x)\}.\]
A subset ${\cal L}(X)$ of ${\cal C}(X)$ is called a lattice when, given any $f,g\in {\cal L}(X)$, we have
\[f\wedge g\in {\cal L}(X)\mbox{ and }f\vee g\in {\cal L}(X).\]

Proposition. Let ${\cal L}(X)$ be a lattice of continuous real-valued functions defined on a compact space $X$. Suppose that the function $h$ defined by
\[h(x)=\inf_{f\in {\cal L}(X)}f(x)\]
is continuous. Then, given any $\epsilon >0$, there exists $g\in {\cal L}(X)$ satisfying $0\leq g(x)-h(x)<\epsilon$ for all $x\in X$. $\sharp$

\begin{equation}{\label{top252}}\tag{20}\mbox{}\end{equation}
Proposition \ref{top252}. Let ${\cal L}(X)$ be a lattice of continuous real-valued functions defined on a compact space $X$ satisfying the following conditions.

${\cal L}(X)$ separates points; that is, for $x\neq y$, there exists $f\in {\cal L}(X)$ satisfying $f(x)\neq f(y)$;
For $f\in {\cal L}(X)$ and $c\in\mathbb{R}$, we have $c\cdot f\in {\cal L}(X)$ and $c+f\in {\cal L}(X)$.

Then, given any continuous real-valued function $h$ defined on $X$ and any $\epsilon >0$, there exists $g\in {\cal L}(X)$ satisfying $0\leq g(x)-h(x)<\epsilon$ for all $x\in X$. $\sharp$

Lemma. Let ${\cal L}(X)$ be a lattice of continuous real-valued functions defined on a compact space $X$ satisfying the conditions of Proposition \ref{top252}. Then, given any $a,b\in\mathbb{R}$ and any $x,y\in X$ with $x\neq y$, there exists $f\in {\cal L}(X)$ such that $f(x)=a$ and $f(y)=b$. $\sharp$

Lemma. Let ${\cal L}(X)$ be a lattice of continuous real-valued functions defined on a compact space $X$ satisfying the conditions of Proposition \ref{top252}. Let $a,b\in\mathbb{R}$ with $a\leq b$, and let $F$ be a closed subset of $X$ with $p\in F$. Then, there exists $f\in {\cal L}(X)$ satisfying $f(x)\geq a$ for all $x\in X$, $f(p)=a$, and $f(x)>b$ for all $x\in F$. $\sharp$

\begin{equation}{\label{top250}}\tag{21}\mbox{}\end{equation}
Lemma \ref{top250}. Given $\epsilon >0$, there exists a polynomial $P$ in one variable satisfying $|P(s)-|s||<\epsilon$ for all $s\in [-1,1]$. $\sharp$

A vector space ${\cal A}(X)$ of functions in ${\cal C}(X)$ is called an algebra when the product of any two functions in ${\cal A}(X)$ is again in ${\cal A}(X)$. Therefore ${\cal A}(X)$ is an algebra if and only if, for $f,g\in {\cal A}(X)$ and $a,b\in\mathbb{R}$, we have $a\cdot f+b\cdot g\in {\cal A}(X)$ and $f\cdot g\in {\cal A}(X)$. A family ${\cal A}(X)$ of functions defined on $X$ is said to separate points when, given $x,y\in X$ with $x\neq y$, there exists $f\in {\cal A}(X)$ satisfying $f(x)\neq f(y)$.

\begin{equation}{\label{top251}}\tag{22}\mbox{}\end{equation}
Theorem \ref{top251}. (Stone-Weierstrass). Let $X$ be a compact space, and let ${\cal A}(X)$ be an algebra of continuous real-valued functions defined on $X$ such that ${\cal A}(X)$ separates the points of $X$ and contains the constant functions. Then, given any continuous real-valued functions $f$ defined on $X$ and any $\epsilon >0$, there exists $g\in {\cal A}(X)$ such that $|g(x)-f(x)|<\epsilon$ for all $x\in X$. In other words, ${\cal A}(X)$ is a dense subset of ${\cal C}(X)$.

Proof. Let $\mbox{cl}({\cal A}(X))$ denote the closure of ${\cal A}(X)$ that is considered as a subset of ${\cal C}(X)$. Therefore for any $f\in\mbox{cl}({\cal A}(X))$,we see that $f$ is a uniform limit of sequences of functions from ${\cal A}(X)$. It is easy to verify that $\mbox{cl}({\cal A}(X))$ is itself an algebra of continuous real-valued functions defined on $X$. We want to prove that $\mbox{cl}({\cal A}(X))={\cal C}(X)$. This will follow from Proposition \ref{top252} if we can show that $\mbox{cl}({\cal A}(X))$ is a lattice. Let $f\in \mbox{cl}({\cal A}(X))$ with $\parallel f\parallel\leq 1$. Then, given $\epsilon >0$, Lemma \ref{top250} says that there exists a polynomial $P$ satisfying $\parallel |f|-P(f)\parallel <\epsilon$. Since $\mbox{cl}({\cal A}(X))$ is an algebra containing the constant functions, we have $P(f)\in\mbox{cl}({\cal A}(X))$. Since $\mbox{cl}({\cal A}(X))$ is a closed subset of ${\cal C}(X)$, we also have $|f|\in {\cal A}(X)$. For any $f\in {\cal A}(X)$, it follows that $|f|/\parallel f\parallel$ has norm $1$, which says $|f|\in\mbox{cl}({\cal A}(X))$. Therefore $\mbox{cl}({\cal A}(X))$ contains the absolute value of each function which is in $\mbox{cl}({\cal A}(X))$. Since
\[f\vee g=\frac{1}{2}(f+g)+\frac{1}{2}|f-g|\mbox{ and }f\wedge g=\frac{1}{2}(f+g)-\frac{1}{2}|f-g|,\]
we conclude that $\mbox{cl}({\cal A}(X))$ is a lattice. The desired result follows immediately from Proposition \ref{top252}. $\blacksquare$

Corollary. Every continuous function on a closed bounded set $X$ in $\mathbb{R}^{n}$ can be uniformly approximated on $X$ by a polynomial in the coordinates.

Proof. The set of all polynomials in the coordinate functions forms an algebra containing the constants. It separates points, since given two distinct points in $\mathbb{R}^{n}$, one of the coordinate functions takes different values on these points. Hence Theorem \ref{top251} applies. $\blacksquare$

Definition. A family ${\cal F}$ of functions from a topological space $(X,\tau_{X})$ into a metric space $(Y,d_{Y})$ is called equicontinuous at the point $x\in X$ when, given $\epsilon >0$, there exists $O\in\tau_{X}$ with $x\in O$ satisfying $d_{Y}(f(x),f(y))<\epsilon$ for all $y\in O$ and all $f\in {\cal F}$. The family is said to be equicontinuous on $X$ when it is equicontinuous at each point $x$ in $X$. $\sharp$

Let $\{f_{n}\}_{n=1}^{\infty}$ be a sequence from an equicontinuous family of functions. We are going to show that if, for each $x\in X$, there exists a convergent subsequence of $\{f_{n}(x)\}_{n=1}^{\infty}$, then $\{f_{n}\}_{n=1}^{\infty}$ has a subsequence which converges uniformly on each compact subset of $X$.

Lemma. Let $\{f_{n}\}_{n=1}^{\infty}$ be a sequence of functions from a countable set $D$ into a topological space $Y$ such that, for each $x\in D$, the closure of the set $\{f_{n}(x)\}_{n=1}^{\infty}$ is sequentially compact. Then, there exists a subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ which converges for each $x\in D$. $\sharp$

Lemma. Let $\{f_{n}\}_{n=1}^{\infty}$ be an equicontinuous sequence of functions from a topological space $(X,\tau_{X})$ into a complete metric space $(Y,d_{Y})$. If the sequences $\{f_{n}(x)\}_{n=1}^{\infty}$ converge for each point $x$ of a dense subset $D$ of $X$, then $\{f_{n}\}_{n=1}^{\infty}$ converges at each point of $X$, and the limit function is continuous. $\sharp$

Lemma. Let $\{f_{n}\}_{n=1}^{\infty}$ be an equicontinuous sequence of functions from a compact topological space $(X,\tau_{X})$ into a metric space $(Y,d_{Y})$. Suppose that the sequence $\{f_{n}(x)\}_{n=1}^{\infty}$ converges at each point $x$ of $X$ to a function $f(x)$. Then $\{f_{n}\}_{n=1}^{\infty}$ converges to $f$ uniformly on $X$. $\sharp$

Theorem. (Ascoli). Let ${\cal F}$ be an equicontinuous family of functions from a separable space $(X,\tau_{X})$ into a metric space $(Y,d_{Y})$. Let $\{f_{n}\}_{n=1}^{\infty}$ be a sequence in ${\cal F}$ such that, for each $x\in X$, the closure of the set $\{f_{n}(x)\}_{n=1}^{\infty}$ is compact. Then, there exists a subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ which converges pointwise to a continuous function $f$, and the convergence is uniform on each compact subset of $X$. $\sharp$

Corollary. Let ${\cal F}$ be an equicontinuous family of real-valued functions defined on a separable space $(X,\tau_{X})$. Then each sequence $\{f_{n}\}_{n=1}^{\infty}$ in ${\cal F}$ which is bounded at each point of a dense subset has a subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ which converges pointwise to a continuous function, and the convergence is uniform on each compact subset of $X$. $\sharp$

It is always true that a closed subset $A$ of a compact space $X$ is compact, for each net in $A$ has a subnet which converges to a point which belongs to $A$ because $A$ is closed. The converse theorem is false, for if $A$ is a proper nonempty subset of an indiscrete space $X$ (only $X$ and the empty set are open), then $A$ is surely compact but not closed. This cannot happen if $X$ is a Hausdorff space.

\begin{equation}{\label{top258}}\tag{23}\mbox{}\end{equation}
Proposition \ref{top258}. If $A$ is a compact subset of a Hausdorff space $X$ and $x$ is a point of $X\setminus A$, then there are disjoint neighborhoods of $x$ and of $A$. Consequently each compact subset of a Hausdorff space is closed.

Proof. Since $X$ is Hausdorff there is a neighborhood $U$ of each $y$ of $A$ such that $x$ does not belong to the closure $\mbox{cl}(U)$. Because $A$ is compact there is a finite family $U_{0},U_{1},\cdots ,U_{n}$ of open sets covering $A$ such that $x\not\in \mbox{cl}(U_{i})$ for $i=0,1,\cdots ,n$. If $V=\bigcup_{i=0}^{n}U_{i}$, then $A\subseteq V$ and $x\not\in \mbox{cl}(V)$. Consequently, $X\setminus \mbox{cl}(V)$ and $V$ are disjoint neighborhoods of $x$ and $A$. $\blacksquare$

Proposition. Let $f$ be a continuous function carrying the compact topological space $X$ onto the topological space $Y$. Then $Y$ is compact, and if $Y$ is Hausdorff and $f$ is one-to-one then $f$ is a homeomorphism.

Proof. If ${\cal A}$ is an open cover of $Y$, then the family of all sets of the form $f^{-1}(A)$ for $A\in {\cal A}$ is an open cover of $X$ which has a finite subcover. The family of images of members of the subcover is a finite subfamily of ${\cal A}$ which covers $Y$ and consequently $Y$ is compact. Suppose that $Y$ is Hausdorff and $f$ is one-to-one. If $A$ is closed subset of $X$, then $A$ is compact and hence its image $f(A)$ is compact and therefore closed. Then $(f^{-1})^{-1}(A)$ is closed for each closed set $A$ and $f^{-1}$ is continuous. $\blacksquare$

\begin{equation}{\label{top259}}\tag{24}\mbox{}\end{equation}
Proposition \ref{top259}. If $A$ and $B$ are disjoint compact subset of a Hausdorff space $X$, then there are disjoint neighborhoods of $A$ and $B$. Consequently, each compact Hausdorff space is normal.

Proof. For each $x\in A$, there is, by Proposition \ref{top258}, a neighborhood of $x$ and a neighborhood of $B$ which are disjoint. Consequently, there is a neighborhood $U$ of $x$ whose closure is disjoint from $B$, and since $A$ is compact there is a finite family $U_{0},U_{1},\cdots ,U_{n}$ such that $\mbox{cl}(U_{i})$ is disjoint from $B$ for $i=0,1,\cdots ,n$ and $A\subseteq V=\bigcup_{i=0}^{n}U_{i}$. Then $V$ is a neighborhood of $A$ and $X\setminus\mbox{cl}(V)$ is a neighborhood of $B$ which is disjoint from $V$. $\blacksquare$

Proposition. If $X$ is a regular topological space, $A$ a compact subset, and $U$ a neighborhood of $A$, then there is a closed neighborhood $V$ of $A$
satisfying $V\subseteq U$. Consequently each compact regular space is normal.

Proof. Because $X$ is regular, for each $x\in A$, there is an open neighborhood $W$ of $x$ such that $\mbox{cl}(W)\subseteq U$, and by compactness there is a finite open cover $W_{0},W_{1},\cdots ,W_{n}$ of $A$ such that $\mbox{cl}(W_{i})\subseteq U$ for each $i$. Then $V=\bigcup_{i=0}^{n}\mbox{cl}(W_{i})$ is the required neighborhood of $A$. $\blacksquare$

Proposition. If $X$ is a completely regular space, $A$ is a compact subset and $U$ is a neighborhood of $A$, then there is a continuous function $f$ on $X$ to the closed interval $[0,1]$ such that $f$ is one on $A$ and zero on $X\setminus A$.

Proof. For each $x\in A$ there is a continuous function $g$ which is one at $x$ and zero on $X\setminus U$. The set $\{y:g(y)>1/2\}$ is open in $X$ and
hence if $h$ is defined by $h(y)=\min\{2g(y),1\}$, then $h$ is continuous, has values in $[0,1]$, is zero on $X\setminus U$, and is one on a neighborhood of $x$. Because $A$ is compact there is a finite family $h_{0},h_{1},\cdots ,h_{n}$ of continuous functions on $X$ to $[0,1]$ satisfying $A\subseteq\bigcup_{i=0}^{n}f^{-1}(\{1\})$ and each $h_{i}$ is zero on $X\setminus U$. The function $f$ whose value at $x$ is $\max_{i}h_{i}(x)$
is the required function. $\blacksquare$

Most of the results of this section are easy consequences of the following theorem.

Theorem. (Wallace). If $X$ and $Y$ are topological spaces, $A$ and $B$ are compact subsets of $X$ and $Y$ respectively, and $W$ is a neighborhood of $A\times B$ in the product space $X\times Y$, then there are neighborhoods $U$ of $A$ and $V$ of $B$ such that $U\times V\subseteq W$.

Proof. For each member $(x,y)$ of $A\times B$ there are open neighborhoods $R$ of $x$ and $S$ of $y$ such that $R\times S\subseteq W$. Since $B$ is compact, for a fixed $x\in A$ there are neighborhoods $R_{i}$ of $x$ and corresponding open sets $S_{i}$ for $i-0,1,\cdots ,n$ such that $B\subseteq Q=\bigcup_{i=0}^{n}S_{i}$. If $P=\bigcap_{i=0}^{n}R_{i}$, then $P$ is a neighborhood of $x$ and $Q$ is a neighborhood of $B$ such that $P\times Q\subseteq W$. From the above concluson, since $A$ is compact there are open sets $P_{i}$ in $X$ and $Q_{i}$ in $Y$, for $i=1,\cdots ,m$ such that each $Q_{i}$ is a neighborhood of $B$, $P_{i}\times Q_{i}\subseteq W$, and $A\subseteq\bigcup_{i=0}^{m}=U$. Then $U$ and $V=\bigcap_{i=0}^{m}Q_{i}$ are neighborhoods of $A$ and $B$ respectively, $U\times V$ is a subset of $W$, and this completes the proof. $\blacksquare$

\begin{equation}{\label{top85}}\tag{25}\mbox{}\end{equation}
Theorem \ref{top85}. (Tychonoff). Let $\prod_{i\in\mathbb{N}}X_{i}$ be the product space of the countable family of nonempty topological spaces $(X_{i},\tau_{i})$ for $i\in\mathbb{N}$. Then $\prod_{i\in\mathbb{N}}X_{i}$ is compact in the product topology if and only if each component space is compact.

Proof. Suppose that $\prod_{i\in\mathbb{N}}X_{i}$ is compact. Since the projection mapping $p_{j}:\prod_{i\in\mathbb{N}}X_{i}\rightarrow X_{j}$ is continuous and onto for each $j\in\mathbb{N}$, Proposition \ref{top61} says that each $X_{j}$ is compact. For the converse, suppose that each $X_{i}$ is compact. Let $\{{\bf x}_{\alpha}\}_{\alpha\in\Lambda}$ be any ultranet in the product space $\prod_{i\in\mathbb{N}}X_{i}$, where the $j$th component of ${\bf x}_{\alpha}$ is denoted by $x_{\alpha}^{(j)}$. Proposition \ref{top62} says that
\[\{p_{j}\left ({\bf x}_{\alpha}\right )\}_{\alpha\in\Lambda}=\left\{x_{\alpha}^{(j)}\right\}_{\alpha\in\Lambda}\]
is an ultranet in $X_{j}$. Proposition \ref{top63} says that the ultranet $\{x_{\alpha}^{(j)}\}_{\alpha\in\Lambda}$ is convergent. From Proposition 56 in page topological spaces, it follows that the ultranet $\{{\bf x}_{\alpha}\}_{\alpha\in\Lambda}$ is also convergent in the product space $\prod_{i\in\mathbb{N}}X_{i}$. Using Proposition \ref{top63} again, we conclude that the product space is compact. This completes the proof. $\blacksquare$

\begin{equation}{\label{top72}}\tag{26}\mbox{}\end{equation}
Theorem \ref{top72}. (Tychonoff). Let $\{(X_{\alpha},\tau_{\alpha})\}_{\alpha\in\Lambda}$ be a family of compact topological spaces. Then, the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$ is compact in the product topology.

Proof. Let $p_{\beta}$ be the projection from the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$ into $X_{\beta}$ for $\beta\in\Lambda$. Then, the following set
\[{\cal S}=\left\{p_{\beta}^{-1}(O):O\in\tau_{\beta}\mbox{ for }\beta\in\Lambda\right\}\]
forms a subbase for the product topology. Let ${\cal A}$ be any collection of members of ${\cal S}$ such that ${\cal A}$ does not contain a finite subcover of the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$. Let
\[{\cal A}_{\beta}=\left\{O:O\in\tau_{\beta}\mbox{ and }p_{\beta}^{-1}(O)\in {\cal A}\right\}.\]
If ${\cal A}_{\beta}$ contains a finite subcover of $X_{\beta}$, then $\{p_{\beta}^{-1}(O)\}_{O\in {\cal A}_{\beta}}\subseteq {\cal A}$ also contains a finite subcover of the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$. This contradiction says that ${\cal A}_{\beta}$ cannot contain a finite subcover of $X_{\beta}$. Since $X_{\beta}$ is compact, it follows that ${\cal A}_{\beta}$ cannot cover $X_{\beta}$. Therefore, for each $\beta\in\Lambda$, we can choose
\[x_{\beta}\in X_{\beta}\setminus\bigcup_{A\in {\cal A}_{\beta}}A.\]
Then, we can form an element ${\bf x}$ in $\prod_{\alpha\in\Lambda}X_{\alpha}$. We also see that ${\bf x}$ is not in the union of members of ${\cal A}$; that is, ${\cal A}$ cannot cover the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$. In other words, any collection of members of the subbase which covers the product space $\prod_{\alpha\in\Lambda}X_{\alpha}$ contains a finite subcover of $\prod_{\alpha\in\Lambda}X_{\alpha}$. Proposition \ref{top66} says that $\prod_{\alpha\in\Lambda}X_{\alpha}$ is compact, and the proof is complete. $\blacksquare$

Theorem. (Tychonoff). The Cartesian product of a collection of compact topological spaces is compact relative to the product topology.

Proof. Let $Q=\prod_{a\in A}X_{a}$ where each $X_{a}$ is a compact topological space and $Q$ has the product topology. Let ${\cal S}$ be the subbase for the product topology consisting of all sets of the form $P_{a}^{-1}(U)$ where $P_{a}$ is the projection into the $a$-th coordinate space and $U$ is open in $X_{a}$. In view of Theorem \ref{top257}, the space $Q$ will be compact if each subfamily ${\cal A}$ of ${\cal S}$, such that no finite subfamily of ${\cal A}$ covers $Q$, fails to cover $Q$. For each index $a$, let ${\cal B}_{a}$ be the family of all open sets $U$ in $X_{a}$ satisfying $P_{a}^{-1}(U)\in {\cal A}$. Then no finite subfamily of ${\cal B}_{a}$ covers $X_{a}$ and hence by compactness there is a point $x_{a}$ satisfying $x_{a}\in X_{a}\setminus U$ for each $U\in {\cal B}_{a}$. The point $x$ whose $a$-th coordinate is $x_{a}$ then belongs to no member of ${\cal A}$ and consequently ${\cal A}$ is not a cover. $\blacksquare$

A subset of a pseudo-metric space is bounded when it is of finite diameter.

Theorem. (Heine-Borel). A subset of Euclidean $n$-space is compact if and only if it is closed and bounded. $\sharp$

The closed unit interval is compact and consequently each cube (the product of closed unit intervals) is compact.

Proposition. A topological space is a Tychonoff space if and only if it is homeomorphic to a subspace of a compact Hausdorff space.

Proof. By Proposition \ref{top263}, each Tychnoff space is homeomorphic to a subset of a cube, which is a compact Hausdorff space. Conversely, each compact Hausdorff space is normal and consequently (Urysohn’s Lemma) is a Tychnoff space, and each subspace is therefore a Tychnoff space. $\blacksquare$

A set in a topological space is {\bf nowhere dense} in the space if and only if its closure has a empty interior.

Proposition. If an infinite number of the coordinate spaces are non-compact, then each compact subset of the product is nowhere dense. $\sharp$

There is an extremely useful lemma of Lebesgue which states that, if ${\cal U}$ is an open cover of a closed interval of real numbers, then there is a positive number $r$ such that, if $|x-y|<r$, then both $x$ and $y$ belong to some member of the cover.

Proposition. If ${\cal U}$ is an open cover of a compact subset $A$ of a pseudo-metric space $(X,d)$, then there is a positive number $r$ such that the open $r$-sphere about each point of $A$ is contained in some member of ${\cal U}$. $\sharp$

There is a useful corollary of the foregoing proposition. If $A$ is a compact subset of a pseudo-metric space and $U$ is a neighborhood of $A$, then there is a positive number $r$ such that $U$ contains the open $r$-sphere about every point of $A$; that is, the distance from $A$ to $X\setminus U$ is positive. Also the foregoing proposition may be rephrased in a suggestive way. If $V$ is the set of all pairs of points of $X$ satisfying $d(x,y)<r$, then $V(x)=\{y:(x,y)\in V\}$ is simply the open $r$-sphere about $x$. The set $V$ is an open subset of $X\times X$ and contains the diagonal $\nabla$ (the set of all pairs $(x,x)$ for $x$ in $X$). The foregoing proposition then implies the following topological result: If ${\cal U}$ is an open cover of a compact pseudo-metric space, then
there is a neighborhood $V$ of the diagonal in $X\times X$ such that each point $x$ the set $V(x)$ is contained in some member of ${\cal U}$.

A cover ${\cal U}$ of a topological space is called an even cover when there is a neighborhood $V$ of the diagonal in $X\times X$ such that for each $x$ the set $V(x)$ is contained in some member of ${\cal U}$. In other words, the family of all sets of the form $V(x)$ defines ${\cal U}$. Recall that a cover ${\cal A}$ is a refinement of ${\cal U}$ if and only if each member of ${\cal A}$ is s subset of some member of ${\cal U}$, and that a family ${\cal B}$ of sets is locally finite if and only if there is a neighborhood of each point of the space which intersects only finitely many members of ${\cal B}$. A family of sets is closed if and only if each member is closed.

Proposition. If an open cover of a space has a closed locally finite refinement then it is an even cover. Consequently each open cover of a compact regular
space is even. $\sharp$

Definition. A topological space $X$ is said to be countably compact when every countable open covering has a finite subcovering. $\sharp$

Any compact space is clearly countably compact. Since any open cover of a Lindel\”{o}f space has a countable subcover, it follows that a countably compact Lindel\”{o}f space is compact.

Proposition. The continuous image of a countably compact space is countably compact. $\sharp$

Definition. A topological space $X$ is said to have the Bolzano-Weieratrass property when every sequence $\{x_{n}\}_{n=1}^{\infty}$ in $X$ has at least one cluster point. $\sharp$

Equivalently, a topological space $X$ has the Bolzano-Weieratrass property if and only if there exists $x\in X$ such that, for each open set $O$ containing $x$ and for each integer $n_{0}$, there exists $n\geq n_{0}$ satisfying $x_{n}\in O$.

\begin{equation}{\label{royp97}}\tag{27}\mbox{}\end{equation}
Proposition \ref{royp97}. A topological space has the Bolzano-Weierstrass property if and only if it is countably compact.

Proof. We first observe that $X$ is countably compact if and only if every countable family ${\cal F}$ of closed sets with the finite intersection property has a nonempty intersection. Suppose now that $X$ has the Bolzano-Weierstrass property and that ${\cal F}=\{F_{i}\}$ is a countable family of closed sets with the finite intersection property. Since the intersection $H_{n}=\bigcap_{k=1}^{n}F_{k}$ is nonempty for every $n$, we may choose an element $x_{n}\in H_{n}$ for each $n$. By the Bolzano-Weierstrass property, the sequence $\{x_{n}\}$ has a cluster point $x$. But $x_{n}\in F_{i}$ for all $n\geq i$ since $x_{n}\in H_{n}$, and so $x$ must belong to $F_{i}$ since $F_{i}$ is closed. Thus $x$ belongs to every $F_{i}$ and so to their intersection. Suppose, on the other hand, that $X$ is countably compact and that $\{x_{i}\}$ is a sequence from $X$. Let $B_{n}$ be the set $\{x_{n},x_{n+1},\cdots\}$. Then $\{\mbox{cl}(B_{n})\}$ is a countable collection of closed sets with the finite intersection property, and so there is a point $x\in\bigcap \mbox{cl}(B_{n})$. This point $x$ is a cluster point of the sequence, since given $N$ and any open set $O$ containing $x$ we have $x\in \mbox{cl}(B_{N})$, and so there must be an $x_{n}\in O$ with $n\geq N$. $\blacksquare$

Definition. A topological space $X$ is said to be sequentially compact when every sequence in $X$ has a convergent subsequence. $\sharp$

\begin{equation}{\label{royp98}}\tag{28}\mbox{}\end{equation}
Proposition \ref{royp98}. A sequentially compact set is countably compact. A countably compact set satisfying the first axiom of countability is sequentially compact. $\sharp$

\begin{equation}{\label{royp99}}\tag{29}\mbox{}\end{equation}
Proposition \ref{royp99}. Let $f$ be a continuous real-valued function defined on a countably compact space $X$. Then $f$ is bounded and assumes its maximum and minimum. $\sharp$

A real-valued function $f$ defined on a topological space is called upper semi-continuous when, for each real number $\alpha$, the set $\{x:f(x)<\alpha\}$ is open. If $f$ is continuous, both $f$ and $-f$ are upper semicontinuous. This implies that Proposition \ref{royp99} is a corollary of the following proposition.

Proposition. Let $f$ be an upper semicontinuous real-valued function defined on a countably compact space $X$. Then $f$ is bounded from above and assumes its maximum.

Proof. The sets $O_{n}=\{x:f(x)<n\}$ form a countable open covering for $X$ and so there must be a finite subcovering $\{O_{1},\cdots ,O_{N}\}$. But this implies $X\subseteq O_{N}$ since $O_{n}\subseteq O_{m}$ for $n<m$. Hence $f(x)<N$ for all $x\in X$ and $f$ is bounded from above. Let $\beta =\sup\{f(x):x\in X\}$. Then, the sets $F_{n}=\{x:f(x)\geq\beta -1/n\}$ form a countable collection of closed sets with the finite intersection property. Hence there is a $y$ belong to every $F_{n}$. Then $f(y)\geq\beta -1/n$ for all $n$; that is, $f(y)\geq\beta$, and also $f(y)=\beta$. $f$ assumes its maximum at $y$. $\blacksquare$

Proposition. (Dini). Let $\{f_{n}\}_{n=1}^{\infty}$ be a sequence of upper semicontinuous real-valued functions defined on a countably compact space $X$, and suppose that for each $x\in X$ the sequence $\{f_{n}(x)\}_{n=1}^{\infty}$ decreases monotonically to zero. Then $\{f_{n}\}_{n=1}^{\infty}$ converges to zero uniformly.

Proof. Choose $\epsilon >0$, and let $O_{n}=\{x:f_{n}(x)<\epsilon\}$. Since $f_{n}$ is upper semicontinuous, $O_{n}$ is open. Since $f_{n}(x)\rightarrow 0$ for each $x$, we have $X\subseteq\bigcup O_{n}$. By the countable compactness of $X$, there are a finite number of open sets $\{O_{1},\cdots ,O_{N}\}$ whose union contains $X$. But this implies $O_{N}=X$, and hence $f_{N}(x)<\epsilon$ for all $x$. If $n\geq N$, we have \[0\leq f_{n}(x)\leq f_{N}(x)<\epsilon,\] and the sequence $\{f_{n}\}$ converges to $0$ uniformly. $\blacksquare$

Definition. A topological space is locally compact when each point has at least one compact neighborhood. $\sharp$

It is obvious that a compact space $X$ is locally compact, since $X$ is a compact neighborhood of any $x\in X$.

\begin{equation}{\label{top80}}\tag{30}\mbox{}\end{equation}
Proposition \ref{top80}. Suppose that the topological space $(X,\tau )$ is locally compact. Then, given any $x\in X$, there exists a compact set $K$ satisfying $x\in\mbox{int}(K)$.

Proof. By definition, there exists a compact neighborhood $N$ of $x$ satisfying $x\in O\subseteq N$, where $O$ is an open set. Since
\[O=\mbox{int}(O)\subseteq\mbox{int}(N),\]
by taking $K=N$, we complete the proof. $\blacksquare$

\begin{equation}{\label{top81}}\tag{31}\mbox{}\end{equation}
Lemma \ref{top81}. Let $(X,\tau )$ be a Hausdorff space. Suppose that, given any $x\in X$, there exists a compact set $K$ satisfying $x\in\mbox{int}(K)$. Then, given any $x\in X$ and any neighborhood $N$ of $x$, there exists an open set $O$ such that $\mbox{\em cl}(O)$ is compact and
\[x\in O\subseteq\mbox{cl}(O)\subseteq N.\]

Proof. We consider the compact subspace $(K,\tau_{K})$ of $(X,\tau )$. The assumption says that, given any $x\in X$, there exists a $\tau$-compact set $K$ satisfying $x\in\tau\mbox{-int}(K)$. Proposition \ref{top68} says that the subspace $(K,\tau_{K})$ is also a Hausdorff space. Given any $\tau$-neighborhood $N$ of $x$, we see that $N^{\circ}\equiv N\cap\tau\mbox{-int}(K)$ is also a $\tau$-neighborhood of $x$. Therefore, there exists a $\tau$-open set $O^{\circ}$ satisfying $x\in O^{\circ}\subseteq N^{\circ}$. Since $K$ is also $\tau_{K}$-compact by Proposition \ref{top73}, it follows that $K$ is $T_{3}$-space by Proposition \ref{top69}. Since
\[O^{\circ}\subseteq N^{\circ}\subseteq\tau\mbox{-int}(K)\subseteq K,\]
we have $O^{\circ}\cap K=O^{\circ}$, which also says that $O^{\circ}$ is $\tau_{K}$-open. Proposition 44 in page topological spaces says that there exists a $\tau_{K}$-open $O$ satisfying
\begin{equation}{\label{top70}}\tag{32}
x\in O\subseteq\tau_{K}\mbox{-cl}(O)\subseteq O^{\circ}\subseteq N^{\circ}\subseteq\tau\mbox{-int}(K).
\end{equation}
Since $O\subseteq \tau\mbox{-int}(K)$, Proposition 14 in page topological spaces says that $O$ is also $\tau$-open. From Proposition \ref{top59}, we also see that $K$ is $\tau$-closed. Therefore, using Proposition 15 in page topological spaces, we obtain
\[\tau\mbox{-cl}(O)=\tau_{K}\mbox{-cl}(O)\subseteq K.\]
Proposition \ref{top57} says that $\tau\mbox{-cl}(O)$ is $\tau$-compact. Using (\ref{top70}), we have
\[x\in O=\tau\mbox{-int}(O)\subseteq\tau\mbox{-cl}(O)\subseteq N^{\circ}\subseteq N.\]
This shows that $X$ is locally compact, and the proof is complete. $\blacksquare$

\begin{equation}{\label{top83}}\tag{33}\mbox{}\end{equation}
Proposition \ref{top83}. Let $(X,\tau )$ be a Hausdorff space. Then $(X,\tau )$ is locally compact if and only if for each $x\in X$, there exists an open set $O$ containing $x$ such that $\mbox{cl}(O)$ is compact.

Proof. Suppose that $(X,\tau )$ is locally compact. Given any $x\in X$, there exists a compact neighborhood $N$ of $x$, i.e., there exists an open set $O$ satisfying $x\in O\subseteq N$. Since $N$ is also closed by Proposition \ref{top59}, it follows $\mbox{cl}(O)\subseteq N$. Proposition \ref{top57} says that $\mbox{cl}(O)$ is compact. The converse is obvious. This completes the proof. $\blacksquare$

Proposition. Any locally compact Hausdorff space is a $T_{3}$-space.

Proof. From Proposition \ref{top80} and Lemma \ref{top81}, given any $x\in X$ and any open set $O$ containing $x$, there exists an open set $O^{*}$ satisfying
\[x\in O^{*}\subseteq\mbox{cl}(O^{*})\subseteq O.\]
Using Proposition 40 in page topological spaces, we see that $X$ is a $T_{3}$-space. This completes the proof. $\blacksquare$

Proposition. Every closed subspace of a locally compact space is locally compact.

Proof. Let $(Y,\tau_{Y})$ be a closed subspace of the locally compact space $(X,\tau_{X})$. Given any $x\in Y$, there exists a $\tau_{X}$-compact neighborhood $N$ of $x$. Proposition \ref{top82} says that $Y\cap N$ is also $\tau_{Y}$-compact. Since $Y\cap N$ is also a $\tau_{Y}$-neighborhood of $x$ by Proposition 16 in page topological spaces, it follows that $N\cap Y$ is a $\tau_{Y}$-compact neighborhood of $x$. This completes the proof. $\blacksquare$

Definition. A function $f$ from a topological space $(X,\tau_{X})$ into a topological space $(Y,\tau_{Y})$ is said to be an open mapping when, given any $\tau_{X}$ open set $O$, the image $f(O)$ is a $\tau_{Y}$-open set. $\sharp$

\begin{equation}{\label{top84}}\tag{34}\mbox{}\end{equation}
Proposition \ref{top84}. Suppose that the function $f:(X,\tau_{X})\rightarrow (Y,\tau_{Y})$ is a continuous, surjective and open mapping. If $X$ is locally compact, $Y$ is also locally compact.

Proof. Given any $y\in Y$, there exists $x\in X$ satisfying $f(x)=y$. The local compactness for $\tau_{X}$ says that there exists a $\tau_{X}$-compact neighborhood $N$ of $x$ satisfying $x\in O\subseteq N$, where $O$ is $\tau_{X}$-open. Then, we have
\[y=f(x)\in f(O)\subseteq f(N).\]
Since $f(N)$ is $\tau_{Y}$-compact by Proposition \ref{top61} and $f(O)$ is $\tau_{Y}$-open by the assumption of open mapping, it follows that $f(N)$ is a $\tau_{Y}$-compact neighborhood of $y$. This shows that $Y$ is locally compact. $\blacksquare$

Proposition. Let $\prod_{i=1}^{\infty}X_{i}$ be the product space of the countable family of topological spaces $(X_{i},\tau_{i})$ for $i=1,2,\cdots$. Then $\prod_{i=1}^{\infty}X_{i}$ is locally compact if and only if each component space is locally compact and all except a finite number of component spaces are compact.

Proof. Suppose that the product space $\prod_{i=1}^{\infty}X_{i}$ is locally compact. Then, the projection $p_{j}:\prod_{i=1}^{\infty}X_{i}\rightarrow X_{j}$ is a continuous, surjective and open mapping for each $j=1,2\cdots$. Proposition \ref{top84} says that each $X_{j}$ is locally compact. We need to show that all except a finite number of component spaces are compact. Given any ${\bf x}\in\prod_{i=1}^{\infty}X_{i}$, there exists a compact neighborhood $N_{\bf x}$ of ${\bf x}$ satisfying
\[{\bf x}\in\prod_{i=1}^{\infty}O_{i}\subseteq N_{\bf x},\]
where $O_{i}=X_{i}$ for all except a finite number $i$ with $O_{i}\in\tau_{i}$. Therefore $p_{j}(N_{\bf x})=X_{j}$ for all except a finite number $j$. Since $p_{j}$ is continuous and $N_{\bf x}$ is compact, by Proposition \ref{top61}. it follows that $X_{j}$ is compact for all except a finite number $j$.

For the converse, let ${\bf x}\in\prod_{i=1}^{\infty}X_{i}$, and let $x_{i}$ be the $i$th coordinate of ${\bf x}$. Since $X_{i_{1}},\cdots ,X_{i_{n}}$ are locally compact for finitely many $i_{k}$ with $k=1,\cdots ,n$, there exists a $\tau_{i_{k}}$-compact neighborhood $N_{x_{i_{k}}}$ of $x_{i_{k}}$ satisfying $x_{i_{k}}\in O_{i_{k}}\subseteq N_{x_{i_{k}}}$. This also says that each $X_{i}$ is compact for $i\not\in\{i_{1},\cdots ,i_{n}\}$. Let $\tau$ denote the product topology for the product space $\prod_{i=1}^{\infty}X_{i}$. We see that $\prod_{i=1}^{\infty}O_{i}$ for $O_{i}=X_{i}$ with $i\not\in\{i_{1},\cdots ,i_{n}\}$ is $\tau$-open in the product space $\prod_{i=1}^{\infty}X_{i}$. Let $\prod_{i=1}^{n}N_{x_{i}}$ for $N_{x_{i}}=X_{i}$ with $i\not\in\{i_{1},\cdots ,i_{n}\}$. We also have
\[{\bf x}\in\prod_{i=1}^{\infty}O_{i}\subset\prod_{i=1}^{\infty}N_{x_{i}}.\]
Since each $N_{x_{i}}$ is $\tau_{i}$-compact, using the Tychonoff’s Theorem \ref{top85}, it follows that $\prod_{i=1}^{n}N_{x_{i}}$ is a $\tau$-compact neighborhood of ${\bf x}$. This shows that the product space is locally compact, and the proof is complete. $\blacksquare$

If $X$ is a locally compact Hausdorff space, we can form a new space $\hat{X}$ by adding to $X$ a single point $\omega\not\in X$ and taking a set in $\hat{X}$ to be open if and only if it is either an open subset of $X$ or the complement of a compact subset in $X$. Then $\hat{X}$ is a compact Hausdorff space. The space $\hat{X}$ is called the Alexandroff one-point compactification of $X$, and $\omega$ is referred to as the point at infinity in $\hat{X}$.

Proposition. Let $K$ be a compact subset of a locally compact Hausdorff space $X$. Then, there is an open set $O$ containing $K$ with $\mbox{cl}(O)$ compact. Given such a set $O$, there is a continuous nonnegative function $f$ on $X$ which vanishes outside $O$ and is identically $1$ on $K$. $\sharp$

Proposition. Let $K$ be a compact subset of a locally compact Hausdorff space and $\{O_{\alpha}\}$ an open covering for $K$. Then, there are a finite number of nonnegative continuous functions $f_{1},\cdots ,f_{n}$ on $X$, each $f_{i}$ vanishing outside a compact set and outside some $O_{\alpha_{i}}$ such that $f_{1}+\cdots f_{n}$ is identically $1$ on $K$. $\sharp$

Proposition. If $X$ is a locally compact topological space which is either Hausdorff or regular, then the family of closed compact neighborhoods of each point is a base for its neighborhood system.

Proof. Let $x$ be a point of $X$, $C$ a compact neighborhood of $x$, and $U$ an arbitrary neighborhood of $x$. If $X$ is regular, then there is a closed neighborhood $V$ of $x$ which is a subset of the intersection of $U$ and the inetrior of $C$, and evidently $V$ is closed and compact. If $X$ is Hausdorff and $W$ is the interior of $U\cap C$, then, since $\mbox{cl}(W)$ is a compact Hausdorff space, $W$ contains a closed compact set $V$ which is a neighborhood of $x$ in $\mbox{cl}(W)$ by Proposition \ref{top259}; but $V$ is also a neighborhood of $x$ in $W$ (that is, with respect to the relativized topology for $W$) and is therefore a neighborhood of $x$ in $X$. $\blacksquare$

In particular, it follows that every locally compact Hausdorff space is regular.

Proposition. If $U$ is a neighborhood of a closed compact subset $A$ of a regular locally compact topological space $X$, then there is a closed compact neighborhood $V$ of $A$ such that $A\subseteq V\subseteq U$. Moreover, there is a continuous function $f$ on $X$ to the closed unit interval such that $f$ is zero on $A$ and one on $X\setminus V$. $\sharp$

It follows that each locally compact, regular, topological space is completely regular and each locally compact Hausdorff space is Tychonoff space. If a function is both open and continuous, then the image of a compact neighborhood of a point is a compact neighborhood of the image point, and consequently the image of a locally compact space is locally compact.

Proposition. If a product is locally compact, then each coordinate space is locally compact and all except a finite number of coordinate spaces are compact. $\sharp$

Proposition. Let $X$ be a topological space, let ${\cal D}$ be an upper semicontinuous decomposition of $X$ whose members are compact, and let ${\cal D}$ have the quotient topology. Then ${\cal D}$ is, respectively, Hausdorff, regular, locally compact, or satisfies the second axiom of countability, provided $X$ has the corresponding property. $\sharp$

There is an interesting corollary to this theorem. If $X$ is separable metric and the members of upper semicontinuous decomposition are compact, then the quotient space is Hausdorff, normal, and satisfies the second axiom of countability, and is consequently metrizable.

Definition. A topological space is paracompact when it is regular and each open cover has an open locally finite refinement. $\sharp$

Recall that a family ${\cal A}$ of subsets of a topological space is discrete if and only if there is a neighborhood of each point of the space which intersects at most one member of the family. The family ${\cal A}$ is $\sigma$-discrete ($\sigma$-locally finite) if and only if it is the union of countably many discrete (locally finite) subfamilies.

Proposition. If $X$ is regular and each open cover has a locally finite refinement, then each open cover has a closed locally finite refinement. $\sharp$

If $U$ is a subset of $X\times X$ and $x\in X$, then $U(x)$ is the set of all points $y$ such that $(x,y)\in U$. If $A$ is a subset of $X$, then
\[U(A)=\{y:(x,y)\in U\mbox{ for some $x$ in $A$}\}\] Clearly, $U(A)$ is the union of the sets $U(x)$ for $x$ in $A$. The set $\{(x,y):(y,x)\in U\}$ is denoted by $U^{-1}$, and $U$ is called symmetric if $U=U^{-1}$. The set $U\cap U^{-1}$ is always symmetric. If $U$ and $V$ are subsets of $X\times X$, then $U\circ V$ is the set of all pairs $(x,z)$ such that for some $y$ in $X$ it is true that $(x,y)\in V$ and $(y,z)\in U$. In other words, $(x,z)\in U\circ V$ if and only if $(x,z)\in V^{-1}(y)\times U(y)$ for some $y$, and consequently $U\circ V$ is the union of the sets $V^{-1}(y)\times U(y)$ for $y$ in $X$. In particular, if $V$ is symmetric, then $V\circ V=\bigcup\{V(y)\times V(y):y\in X\}$. Finally, for each subset $A$ of $X$ it is true that $(U\circ V)(A)=U(V(A))$.

Proposition. Let $X$ be a topological space such that each open cover is even. If $U$ is a neighborhood of the diagonal in $X\times X$ then there is a symmetric neighborhood $V$ of the diagonal such that $V\circ V\subseteq U$. $\sharp$

The preceding proposition has the following intuitive content. Let us say two points $x$ and $y$ are at most $U$-distance apart if $(x,y)\in U$. Then, there is $V$ such that, if $x$ and $y$, and $y$ and $z$, are at most $V$-distance apart, then $x$ and $x$ are at most $U$-distance apart.

Proposition. Let $X$ be a topological space such that each open cover is even and let ${\cal A}$ be a locally finite (or a discrete) family of subsets of $X$. Then, there is a neighborhood $V$ of the diagonal in $X\times X$ such that the family of all sets $V(A)$ for $A$ in ${\cal A}$ is locally finite. $\sharp$

Corollary. A paracompact space is normal. $\sharp$

Proposition. If $X$ is a space such that each open cover is even, then every open cover of $X$ has an open $\sigma$-discrete refinement. $\sharp$

Proposition. If each open cover of a space has an open $\sigma$-locally finite refinement, then each open cover has a locally finite refinement. $\sharp$

Theorem. If $X$ is a regular topological space, then the following statements are equivalent.

(a) The space $X$ is paracompact.

(b) Each open cover of $X$ has a locally finite refinement.

(d) Each open cover of $X$ is even.

(e) Each open cover of $X$ has an open $\sigma$-discrete refinement.

(f) Each open cover of $X$ has an open $\sigma$-locally finite refinement. $\sharp$

Proposition \ref{kelp14} states that each open cover of a pseudo-metrizable space has an open $\sigma$-discrete refinement. This fact and the foregoing theorem give the following corollary.

Corollary. Each pseudo-metrizable space is paracompact. $\sharp$

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

Equivalent Compactness in Metric Spaces.

We shall show that in a metric space the notions of compactness, countable compactness, and sequential compactness are equivalent.

\begin{equation}{\label{royl912}}\tag{35}\mbox{}\end{equation}
Lemma \ref{royl912}. Let $X$ be a countably compact metric space. Then, given $\epsilon >0$, there are a finite number of points $x_{1},\cdots ,x_{n}$ in $X$ such that, for each $x\in X$, there exists $x_{k}$ satisfying $d(x,x_{k})<\epsilon$.

Proof. Suppose that no such finite collection of points exists. Then, we can choose an infinite sequence $\{x_{n}\}_{n=1}^{\infty}$ of points in $X$ satisfying $d(x_{n},x_{m})\geq\epsilon$ for $n\neq m$. Since each spheroid of radius $\epsilon /3$ can contain at most one term of this sequence, the Bolzano-Weierstrass property does not hold, and $X$ is not countably compact. This completes the proof. $\blacksquare$

\begin{equation}{\label{royp913}}\tag{36}\mbox{}\end{equation}
Proposition \ref{royp913}. A countably compact metric space is separable.

Proof. Using Lemma \ref{royl912}, for each positive integer $n$, let $F_{n}$ be a finite set of points such that, for each $x\in X$, there exists $y\in F_{n}$ satisfying $d(x,y)<1/n$. Then $\bigcup_{n=1}^{\infty}F_{n}$ is a countable dense subset of $X$. This completes the proof. $\blacksquare$

\begin{equation}{\label{royc914}}\tag{37}\mbox{}\end{equation}
Corollary \ref{royc914}. For a metric space the notions of compactness, countable compactness, and sequential compactness are equivalent.

Proof. Since a metric space satisfies the first axiom of countability, the notions of countable compactness and sequential compactness coincide by Proposition \ref{royp98}. Compactness trivially implies countable compactness. If $X$ is countably compact, it must satisfy the second axiom of countability by Proposition \ref{royp913}, and so be compact. $\blacksquare$

\begin{equation}{\label{krel252}}\tag{38}\mbox{}\end{equation}
Proposition \ref{krel252}. A compact subset $M$ of a metric space is closed and bounded.

Proof. For every $x\in \mbox{cl}(M)$, there exists a sequence $\{x_{n}\}_{n=1}^{\infty}$ in $M$ satisfying $x_{n}\rightarrow x$ by part (i) of Proposition \ref{kret146}. Since $M$ is compact, $x\in M$, $x\in M$. Hence $M$ is closed. We prove that $M$ is bounded. If $M$ were unbounded, it would contain an unbounded sequence $\{y_{n}\}$ such that $d(y_{n},b)>n$, where $b$ is any fixed element. This sequence could not have a convergent subsequence since a convergent subsequence must be bounded by Proposition \ref{krel142}. $\blacksquare$

The converse of the above proposition is not true in general. However, we have the following result.

Proposition. In a finite-dimensional normed space $X$, any subset $M\subseteq X$ is compact if and only if $M$ is closed and bounded.

Proof. Compactness implies closedness and boundedness by Proposition \ref{krel252} since the normed space is also a metric space. We now prove the converse. Let $M$ be a closed and bounded. Let $\dim (X)=n$ and $\{e_{1},\cdots ,e_{n}\}$ a basis for $X$. We consider any sequence $\{x_{m}\}$ in $M$. Each $x_{m}$ has a representation

\[x_{m}=\alpha_{1}^{(m)}e_{1}+\cdots +\alpha_{n}^{(m)}e_{m}.\]

Since $M$ is bounded, so is $\{x_{m}\}$, say, $\parallel x_{m}\parallel\leq k$ for all $m$. By Lemma \ref{top144}, we have
\[k\geq\parallel x_{m}\parallel =\left |\!\left |\sum_{j=1}^{n}\alpha_{j}^{(m)}
e_{j}\right |\!\right |\geq c\sum_{j=1}^{n}|\alpha_{j}^{(m)}|\geq c|\alpha_{j}^{(m)}|\]
where $c>0$. hence the sequence of numbers $\{\alpha_{j}^{(m)}\}$, for $j$ fixed, is bounded and, by the Bolzano-Weierstrass theorem, has a point of accumulation $\alpha_{j}$. As in the proof of Lemma \ref{top144} we conclude that $\{x_{m}\}$ has a subsequence $\{z_{m}\}$ which converges to $z=\sum_{j}\alpha_{j}e_{j}$. Since $M$ is closed, $z\in M$. This shows that the arbitrary sequence $\{x_{m}\}$ in $M$ has a subsequence which converges in $M$. Hence $M$ is compact. $\blacksquare$

Definition. A subset $A$ of a metric space $(X,d)$ is totally bounded when, given any $\epsilon >0$, there exists a finite subset $\{x_{1},\cdots ,x_{n}\}$ of $X$ such that the the finite collection of open balls $\{B(x_{i};\epsilon )\}_{i=1}^{n}$ covers $A$. $\sharp$

We have the following observations.

If $A$ is totally bounded, then $\mbox{cl}(A)$ and any subsets of $A$ are all totally bounded.
Every compact metric space is totally bounded.
Every subset of a totally bounded space is totally bounded.

Proposition. Every totally bounded metric space is separable.

Proof. If $(X,d)$ is totally bounded, then, for each $n\in\mathbb{N}$, we pick a finite subset $F_{n}$ of $X$ such that $X=\bigcup_{x\in F_{n}} B(x;\frac{1}{n})$. The set $F=\bigcup_{n=1}^{\infty}F_{n}$ is countable and dense. $\blacksquare$

Proposition. A metric space is compact if and only if it is both complete and totally bounded.

Proof. If $X$ is compact, it is trivially totally bounded. By Proposition \ref{royp97} and Corollary \ref{royc914}, if $\{x_{n}\}$ is a Cauchy sequence in $X$, then $\{x_{n}\}$ must have a cluster point. But a Cauchy sequence which has a cluster point converges to the cluster point. Thus $X$ is complete. Suppose that $X$ is complete and totally bounded. To show that $X$ is compact, it suffices to show that each infinite sequence $\{x_{n}\}$ has a convergent subsequence by Corollary \ref{royc914}. Since $X$ is totally bounded, we may cover $X$ by a finite number of spheres of radius $1$. Among these spheres there must be a sphere $S_{1}$ which contains infinitely many terms of the sequence $\{x_{n}\}$ since the number of spheres is finite. Covering $X$ by a finite number of spheres of radius $1/2$, we can find among them a sphere $S_{2}$ such that $S_{1}\cap S_{2}$ contains infinitely many terms of sequence $\{x_{n}\}$ since the finite number of spheres with radius $1/2$ also covers $S_{1}$. Continuing we obtain a sequence $\{S_{k}\}$ of spheres, $S_{k}$ having radius $1/k$, such that $S_{1}\cap\cdots\cap S_{k}$ contains infinitely many terms of the sequence. Since there are infinitely many terms of the sequence in $S_{1}\cap\cdots\cap S_{k}$, we may choose $n_{k}$ so that $n_{k}>n_{k-1}$ and $x_{n_{k}}\in S_{1}\cap\cdots\cap S_{k}$. Then $\{x_{n_{k}}\}$ is a subsequence of $\{x_{n}\}$, and it must be a Cauchy sequence, since $d(x_{n_{k}},x_{n_{l}})\leq 2/N$ for $k,l\geq N$. Since $X$ is complete, this subsequence converges. $\blacksquare$

Proposition. A metric space is totally bounded if and only if its completion is compact. $\sharp$

Proposition. Let $f$ be a continuous function of a compact metric space $(X,d_{X})$ into a metric space $(Y,d_{Y})$. Then $f$ is uniformly continuous.

Proof. Given $\epsilon >0$ and $x\in X$, there is a $\delta_{x}>0$ such that $d_{X}(x,y)<\delta_{x}$ implies $d_{Y}(f(x),f(y))<\epsilon /2$. Let $O_{x}$ be the spheroid $\{y:d(x,y)<\frac{1}{2}\delta_{x}\}$. Then $\{O_{x}:x\in X\}$ is an open covering of $X$ and so has a finite subcovering $\{O_{x_{1}},\cdots ,O_{x_{n}}\}$. Let $\delta =\frac{1}{2}\min\{\delta_{x_{1}},\cdots ,\delta_{x_{n}}\}$. Then $\delta >0$. Given two points $y$ and $z$ in $X$ such that $d_{X}(y,z)<\delta$, the point $y$ must belong to some $O_{x_{i}}$, and hence $d_{X}(y,x_{i})<\frac{1}{2}\delta_{x_{i}}$. Consequently, we have

\[d_{X}(z,x_{i})\leq d_{X}(z,y)+d_{X}(y,x_{i})<\frac{1}{2}\delta_{x_{i}}+\delta\leq\delta_{x_{i}}.\]

Therefore, we obtain $d_{Y}(f(y),f(x_{i}))<\epsilon /2$ and $d_{Y}(f(z),f(x_{i}))<\epsilon /2$, which implies $d_{Y}(f(z),f(y))<\epsilon$. Therefore, $f$ is uniformly continuous on $X$. $\blacksquare$

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

Compactification.

In studying a non-compact topological space $X$ it is often convenient to construct a space which contains $X$ as a subspace and is itself compact. For example, it is frequently useful to adjoin two points, $+\infty$ and $-\infty$, to the space of real numbers. The resulting space is sometimes called the extended real numbers; it is linearly ordered by agreeing that $+\infty$ is the largest member and $-\infty$ is the smallest. With this ordering (and extension of the usual ordering) it turns out that every nonempty subset of the extended real numbers has both an infinimum and a supremum and the space is compact relative to its order topology. The extended reals are a compactification of the space of real numbers. The simplest sort of compactification of a topological space is made by adjoining a single point. This procedure is familiar in analysis, for in function theory the complex sphere is constructed by adjoining a single point $\infty$, to the Euclidean plane and specifying that the neighborhoods of $\infty$ are the complements of bounded subsets of the plane. This construction can be duplicated for an arbitrary topological space; the clue to the topology to be introduced in the enlarged space is the fact that the complement of an open neighborhood of $\infty$ in the complex sphere is compact. The one point compactification of a topological space $X$ is the set $X^{*}=X\cup\{\infty\}$ with the topology whose members are the open subsets of $X$ and all subsets $U$ of $X^{*}$ such that $X^{*}\setminus U$ is a closed compact subset of $X$.

Theorem. (Alexandroff). The one point compactification $X^{*}$ of a topological space $X$ is compact and $X$ is a subspace. The space $X^{*}$ is
Hausdorff if and only if $X$ is locally compact and Hausdorff. $\sharp$

If $X$ is a compact topological space, then $\infty$ is an isolated point of the one point compactification (that is, $\{\infty\}$ is both open and closed). Conversely, if $\infty$ is an isolated point of $X^{*}$, then $X$ is closed in $X^{*}$ and is therefore compact. It is convenient to allow a topological embedding rather than insist that the original be actually a subspace of the constructed compact space. With this in mind, a compactification of a topological space $X$ is defined to be a pair $(f,Y)$, where $Y$ is a compact topological space and $f$ is a homeomorphism of $X$ onto a dense subspace of $Y$. (To be consistent, the one point compactification of $X$ should be the pair $(i,X^{*})$, where $i$ is the identity function.) A compactification $(f,Y)$ is called Hausdorff if and only if $Y$ is Hausdorff space. A relation is defined on the collection of all compactifications of a space $X$ by agreeing that $(f,Y)\geq (g,Z)$ if and only if there is a continuous map $h$ of $Y$ onto $Z$ such that $h\circ f=g$. Equivalently, $(f,Y)\geq (g,Z)$ if and only if the function $g\circ f^{-1}$ on $f^{-1}(X)$ to $Z$ has a continuous extension $h$ which carries $Y$ onto $Z$. If the function $h$ can be taken to be a homeomorphism, then $(f,Y)$ and $(g,Z)$ are said to be topologically equivalent. In this case both of the relations $(f,Y)\geq (g,Z)$ and $(g,Z)\geq (f,Y)$ hold, for $h^{-1}$ is a continuous map of $Z$ onto $Y$ such that $f=h^{-1}\circ g$.

\begin{equation}{\label{top260}}\tag{39}\mbox{}\end{equation}
Proposition \ref{top260}. The collection of all compactifications of a topological space is partially ordered by $\geq$. If $(f,Y)$ and $(g,Z)$ are Hausdorff compactifications of a space and $(f,Y)\geq (g,Z)\geq (f,Y)$, then $(f,Y)$ and $(g,Z)$ are topologically equivalent. $\sharp$

The smallest compactification of a compact Hausdorff space $X$ is $X$ itself (more precisely $(i,X)$ where $i$ is the identity map on $X$). One would expect that the one point compactification of a non-compact space would be the smallest relative to the ordering $\geq$. If we restrict our attention to Hausdorff compactifications this is actually the case, although it is easy to see that there is generally no compactification which is smaller than every other. On the other hand, if $X$ is a space which has a Hausdorff compactification, then there is a largest compactification which we now construct.

For each topological space $X$ let $F(X)$ be the family of all continuous functions on $X$ to the closed unit interval $Q$. The cube $Q^{F(X)}$ (the product of the unit interval $Q$ taken $F(X)$ times) is compact by the Tychonoff theorem. The evaluation map $e$ carries a member $x$ of $X$ into the member $e(x)$ of $Q^{F(X)}$ whose $f$-th coordinate is $f(x)$ for each $f$ in $F(X)$. Evaluation is a continuous map of $X$ into the cube $Q^{F(X)}$, and if $X$ is a Tychonoff space, then $e$ is a homeomorphism of $X$ onto a suspace of $Q^{F(X)}$. (The embedding Lemma \ref{top262} states these facts explicitly.) The Stone-Cech compactification is the pair $(e,\beta (X))$ where $\beta (X)$ is the closure of $e(X)$ in the cube $Q^{F(X)}$.

Proposition. If $f$ is a function on a set $A$ to a set $B$ and $f^{*}$ is the map of $Q^{B}$ into $Q^{A}$ defined by $f^{*}(y)=y\circ f$ for all $y$ in $Q^{B}$, then $f^{*}$ is continuous. $\sharp$

Observe that the function $f^{*}$ induced by $f$ goes in the direction opposite to that of $f$, in the sense that $f$ carries $A$ into $B$ while $f^{*}$ carries $Q^{B}$ into $Q^{A}$.

\begin{equation}{\label{top261}}\tag{40}\mbox{}\end{equation}
Theorem \ref{top261}. (Stone-Cech). If $X$ is a Tychonoff space and $f$ is a continuous function on $X$ to a compact topological space $Y$, then there is a continuous extension of $f$ which carries the compactification $\beta (X)$ into $Y$. More precisely, if $(e,\beta (X))$ is the Stone-Cech compactification, then $f\circ e^{-1}$ can be extended to a continuous function on $\beta (X)$ to $Y$. $\sharp$

The extension property of the foregoing theorem shows that the Stone-Cech compactification $(e,\beta (X))$ follows every other Hausdorff compactification in the ordering $\geq$ and is therefore the largest such compactification. If $(f,Y)$ has this extension property, then $(f,Y)\geq (e,\beta (X))$ and consequently is topolically equivalent to $(e,\beta (X))$ by Proposition \ref{top260}. Hence, the compactification $(e,\beta (X))$ is characterized (to a topological equivalence) by the extension property of Theorem \ref{top261}.

Compactness.

Equivalent Compactness in Metric Spaces.

Compactification.

Hsien-Chung Wu

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Compactness.

Equivalent Compactness in Metric Spaces.

Compactification.

Hsien-Chung Wu

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