Albert Bierstadt (1830-1902) was an American painter.
Definition. Let \(M\) be a nonempty set, and let \(d:M\times M\rightarrow\mathbb{R}\) be a real-valued function defined on \(M\times M\). We say that \((M,d)\) is a metric space when the following conditions are satisfied: for any points \(x,y,z\in M\),
- \(d(x,x)=0\);
- \(d(x,y)>0\) when \(x\neq y\);
- \(d(x,y)=d(y,x)\);
- \(d(x,y)\leq d(x,z)+d(z,y)\); this is called the triangle inequality.
In this case, the function \(d\) is called the {\bf metric} of space \((M,d)\), and \(d(x,y)\) describes the distance between points \(x\) and \(y\) in \(M\).$\sharp$.
Example. We provide many examples of metric spaces.
(i) Let \(M=\mathbb{R}^{n}\). Then, the triangle inequality is given by
\[\parallel {\bf x}-{\bf y}\parallel\leq\parallel {\bf x}-{\bf z}\parallel
+\parallel {\bf z}-{\bf y}\parallel .\]
Now, we take
\begin{equation}{\label{maeq424}}\tag{1}
d\left ({\bf x},{\bf y}\right )=\parallel {\bf x}-{\bf y}\parallel .
\end{equation}
Then, we can show that the four conditions in Definition 1 are satisfied. This says that the Euclidean space \(\mathbb{R}^{n}\) is a metric space with the metric \(d\) defined in (\ref{maeq424}).
(ii) Let \(M=\mathbb{R}^{n}\) and define \(d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}\) by
\[d({\bf x},{\bf y})=\left |x_{1}-y_{1}\right |+\cdots +\left |x_{n}-y_{n}\right |.\]
Then, we can show that the four conditions in Definition 1 are satisfied.
(iii) Let \(M=\mathbb{R}^{n}\) and define \(d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}\) by
\[d({\bf x},{\bf y})=\max\left\{\left |x_{1}-y_{1}\right |,\cdots ,\left |x_{n}-y_{n}\right |\right\}.\]
Then, we can show that the four conditions in Definition 1 are satisfied.
(iv) Let \(M\) be the unit circle in \(\mathbb{R}^{2}\), i.e.,
\[M=\left\{(x_{1},x_{2}):x_{1}^{2}+x_{2}^{2}=1\right\}.\]
We define \(d({\bf x},{\bf y})\) to be the length of the smaller arc joining the two points \({\bf x}\) and \({\bf y}\) on the unit circle. Then, we can show that the four conditions in Definition 1 are satisfied.
(v) Let \(M\) be the set of all bounded real-valued functions defined on a subset \(S\) of \(\mathbb{R}\). Given any \(f,g\in M\), we define
\[d(f,g)=\sup_{x\in S}\left |f(x)-g(x)\right |.\]
Then \((M,d)\) is a metric space.
(vi) Let \(M\) be the set of all continuous real-valued functions defined on a compact interval \([a,b]\). Given any \(f,g\in M\), we define
\[d(f,g)=\max_{x\in [a,b]}\left |f(x)-g(x)\right |.\]
Then \((M,d)\) is a metric space.
(vii) Let \(M\) be the set of all continuous real-valued functions defined on a compact interval \([a,b]\). Given any \(f,g\in M\), we define
\[d(f,g)=\int_{a}^{b}\left |f(x)-g(x)\right |dx.\]
Then \((M,d)\) is a metric space.
(viii) Let \((M,d)\) be a metric space. We define
\[\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}\mbox{ for }x,y\in M.\]
Then \((M,\rho)\) is also a metric space. \(\sharp\)
The concept of point set topology in \(\mathbb{R}^{n}\) can be extended to the metric space. The main issue is the definition of open ball. Let \((M,d)\) be a metric space. The open ball \(B(x;r)\) with center \(x\in M\) and radius \(r>0\) is defined by
\[B_{M}(x;r)=\left\{y\in M:d(x,y)<r\right\}.\]
In the real line \(M=\mathbb{R}\), we define the metric \(d(x,y)=|x-y|\) for any \(x,y\in\mathbb{R}\), then the open ball \(B_{\mathbb{R}}(0;1)\) is the open interval \((-1,1)\). If we take \(M=[0,1]\) and define the same metric \(d(x,y)=|x-y|\) for any \(x,y\in M\), then the open ball \(B_{M}(0;1)\) is \([0,1)\).
Definition. Let \((M,d)\) be a metric space, and let \(S\) be a subset of \(M\).
- A point \(a\in S\) is called an interior point when there exists \(\epsilon >0\) satisfying \(B_{M}(a;\epsilon )\subseteq S\).
- The set of all interior point of \(S\) is called the interior of \(S\), and is denoted by \(\mbox{int}(S)\).
- The set \(S\) is called {\bf open} in \(M\) when \(S=\mbox{int}(S)\).
- The set \(S\) is called {\bf closed} in \(M\) when \(S^{c}=M\setminus S\) is open in \(M\). \(\sharp\)
It is clear to see that every open ball \(B_{M}(a;r)\) in a metric space \((M,d)\) is an open subset of \(M\).
\begin{equation}{\label{mapex425}}\tag{2}\mbox{}\end{equation}
Example \ref{mapex425}. Let \(M=[0,1]\). Define \(d(x,y)=|x-y|\) for any \(x,y\in M\). Every interval of the form \([0,x)\) or \((x,1]\) for \(0<x<1\) is an open subset of \(M\). However, they are not open sets in \(\mathbb{R}\). \(\sharp\)
\begin{equation}{\label{map426}}\tag{3}\mbox{}\end{equation}
Proposition \ref{map426}. Let \((M,d)\) be a metric space.
(i) The intersection of finite collection of open sets in \(M\) is an open set in \(M\), and the union of any collection of open sets in \(M\) is also an open set in \(M\).
(ii) The union of finite collection of closed sets in \(M\) is a closed set in \(M\), and the intersection of any collection of closed sets in \(M\) is also a closed set in \(M\).
Proof. The same arguments in the proofs of page point set topology in \(\mathbb{R}^{n}\) are still valid for the metric space. \(\blacksquare\)
Let \(S\) be a subset of \(M\). Then \((S,d)\) is called a metric subspace of \((M,d)\). The open ball in \((S,d)\) is defined to be
\[B_{S}(a;r)=B_{M}(a;r)\cap S\]
for some open ball \(B_{M}(a;r)\) in \(M\). Therefore, we can similarly define the concepts of open and closed sets in \(S\) based on the concept of open ball \(B_{S}(a;r)\) in \(S\). If we take \(M=\mathbb{R}\) and \(S=[1,0]\), then Example \ref{mapex425} says that the open sets \([0,x)\) for \(0<x<1\) in \(S\) are not the open sets in \(M\). The relation between open sets in \(M\) and \(S\) is given below.
\begin{equation}{\label{map427}}\tag{4}\mbox{}\end{equation}
Proposition \ref{map427}. Let \((S,d)\) be a metric subspace of \((M,d)\), and let \(T\) be a subset of \(S\). Then \(T\) is open in \(S\) if and only if \(T=A\cap S\) for some open set \(A\) in \(M\).
Proof. Suppose that \(A\) is open in \(M\). Let \(T=A\cap S\). For any \(x\in T\), we have \(x\in A\). Since \(A\) is open, there exists \(\epsilon >0\) satisfying \(B_{M}(x;\epsilon )\subseteq A\). This says that
\begin{align*} & B_{S}(x;r)=B_{M}(x;r)\cap S\\ & \quad\subseteq A\cap S=T,\end{align*}
and show that \(T\) is open in \(S\).
On the other hand, we assume that \(T\) is open in \(S\). For any \(x\in T\), there exists \(\epsilon >0\) satisfying \(B_{S}(x;r_{x})\subseteq T\), where
\[B_{S}(x;r_{x})=B_{M}(x;r_{x})\cap S.\]
We define
\[A=\bigcup_{x\in T}B_{M}(x;r_{x}).\]
Then \(A\) is open in \(M\) by Proposition \ref{map426}. It is not hard to show that \(T=A\cap S\). This completes the proof. \(\blacksquare\)
Proposition. Let \((S,d)\) be a metric subspace of \((M,d)\), and let \(T\) be a subset of \(S\). Then \(T\) is closed in \(S\) if and only if \(T=A\cap S\) for some closed set \(A\) in \(M\).
Proof. Suppose that \(A\) is closed in \(M\). Let \(T=A\cap S\). This says that \(A^{c}\) is open in \(M\). Since we have
\begin{align*}S\setminus T & =S\setminus (A\cap S)\\ & =S\cap A^{c},\end{align*}
Proposition \ref{map427} says that \(S\setminus T\) is open in \(S\). Therefore, we conclude that \(T\) is closed in \(S\).
On the other hand, we assume that \(T\) is closed in \(S\). Then \(S\setminus T\) is open in \(S\). By Proposition \ref{map427} again, there exists an open set \(O\) in \(M\) satisfying \(S\setminus T=S\cap O\). We define \(U=S\setminus T\). Since \(T\subseteq S\), we have
\begin{align*} T & =S\setminus U\\ & =S\setminus (S\cap O)\\ & =S\setminus O\\ & =S\cap O^{c}.\end{align*}
Since \(O^{c}\) is closed in \(M\), this complete the proof. \(\blacksquare\)
Definition. Let \((M,d)\) be a metric space, and let \(S\) be a subset of \(M\).
- A point \(x\in M\) is called an adherent point of \(S\) when every open ball \(B_{M}(x;r)\) contains at least one point of \(S\), i.e., \(B_{M}(x;r)\cap S\neq\emptyset\).
- A point \(x\in M\) is called an accumulation point of \(S\) when \(x\) is an adherent point of \(S\setminus{x}\).
- If \({\bf x}\in S\) and \({\bf x}\) is not an accumulation point of \(S\), then \({\bf x}\) is called an isolated point.
- The set of all adherent points of \(S\) is called the closure of \(S\) and is denoted by \(\mbox{cl}(S)\).
- The set of all accumulation points of \(S\) is called the derived set and is denoted by \(S’\). \(\sharp\)
Proposition. Let \((M,d)\) be a metric space, and let \(S\) be a subset of \(S\). The following statements are equivalent.
- \(S\) is closed in \(M\).
- \(S\) contains all its adherent points.
- \(S\) contains all its accumulation points.
- We have \(S=\mbox{cl}(S)\).
Proof. The same arguments in the proofs of page point set topology in \(\mathbb{R}^{n}\) are still valid for the metric space. \(\blacksquare\)
\begin{equation}{\label{mat437}}\tag{5}\mbox{}\end{equation}
Theorem \ref{mat437}. Let \((M_{1},d_{1})\) and \((M_{2},d_{2})\) be two metric spaces. We consider a function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2})\).
(i) \(f\) is continuous on \(M_{1}\) if and only if, for every open subset \(Y\) in \(M_{2}\), the inverse image \(f^{-1}(Y)\) is open in \(M_{1}\).
(ii) \(f\) is continuous on \(M_{1}\) if and only if, for every closed subset \(Y\) in \(M_{2}\), the inverse image \(f^{-1}(Y)\) is closed in \(M_{1}\).
Proof. To prove part (i), let \(f\) be continuous on \(M_{1}\), and let \(Y\) be open in \(M_{2}\). Given any point \(p\in f^{-1}(Y)\), we shall prove that \(p\) is an interior point of \(f^{-1}(Y)\). Let \(y=f(p)\). Since \(Y\) is open in \(M_{2}\), there exists an open ball \(B_{M_{2}}(y;\epsilon )\) satisfying \(B_{M_{2}}(y;\epsilon )\subseteq Y\). Since \(f\) is continuous at \(p\), there exists
$\delta>0$ satisfying \(f(B_{M_{1}}(p;\delta ))\subseteq B_{M_{2}}(y;\epsilon )\) by Remark \ref{ma91}. Therefore, using Proposition \ref{ma92}, we have
\begin{align*} B_{M_{1}}(p;\delta ) & \subseteq f^{-1}(f(B_{M_{1}}(p;\delta )))\\ & \subseteq f^{-1}(B_{M_{2}}(y;\epsilon ))\\ & \subseteq f^{-1}(Y),\end{align*}
which says that \(p\) is an interior point of \(f^{-1}(Y)\). For the converse, we assume that \(f^{-1}(Y)\) is open in \(M_{2}\) for every open set \(Y\) in \(M_{1}\). Given any \(p\in M_{1}\),
let \(y=f(p)\). We shall prove that \(f\) is continuous at \(p\). Given any \(\epsilon >0\), the open ball \(B_{M_{2}}(y;\epsilon )\) is open in \(M_{2}\), which also says that \(f^{-1}(B_{M_{2}}(y;\epsilon ))\) is open in \(M_{1}\). Since \(p\in f^{-1}(B_{M_{2}}(y;\epsilon ))\), there exists \(\delta>0\) satisfying \(B_{M_{1}}(p;\delta )\subseteq f^{-1}(B_{M_{2}}(y;\epsilon ))\). Therefore, we obtain
\begin{align*} f(B_{M_{1}}(p;\delta )) & \subseteq f(f^{-1}(B_{M_{2}}(y;\epsilon )))\\ & \subseteq B_{M_{2}}(y;\epsilon ),\end{align*}
which also says that \(f\) is continuous at \(p\) by Remark \ref{ma91} again.
To prove part (ii), if \(Y\) is closed in \(M_{2}\), then \(Y^{c}\) is open in \(M_{2}\) and
\begin{align*} f^{-1}(Y^{c}) & =f^{-1}(M_{2}\setminus Y)\\ & =M_{1}\setminus f^{-1}(Y).\end{align*}
The result follows from part (i) immediately, and the proof is complete. \(\blacksquare\)
The Bolzano-Weierstarss theorem, the Cantor intersection theorem, and the covering theorems of Lindel\”{o}f and Heine-Borel given in page point set topology in \(\mathbb{R}^{n}\) are not generally valid in an arbitrary metric space \((M,d)\), since the proofs used some special properties of \(\mathbb{R}^{n}\). This says that further restrictions on \(M\) are required in order to extend these theorems to metric spaces.
Definition. Let \((M,d)\) be a metric space, and let \(S\) be a subset of \(M\).
- A collection \({\cal F}\) of open sets in \(M\) is said to be an open covering of \(S\) when \(S\subseteq\bigcup_{A\in {\cal F}}A\).
- \(S\) is called compact when every open covering of \(S\) contains a finite subcovering of \(S\).
- \(S\) is called bounded when there exists \(\epsilon >0\) and \(a\in M\) satisfying \(S\subseteq B(a;\epsilon )\). \(\sharp\)
\begin{equation}{\label{mat428}}\tag{6}\mbox{}\end{equation}
Theorem \ref{mat428}. Let \(S\) be a compact subset of a metric space \((M,d)\). Then, we have the following properties.
(i) \(S\) is closed and bounded.
(ii) Every infinite subset of \(S\) has an accumulation point of \(S\).
Proof. To prove part (i), the proof of Theorem A in page point set topology in \(\mathbb{R}^{n}\) for proving that statement (a) implies statement (b) is still valid when the Euclidean distance \(\parallel {\bf x}-{\bf y}\parallel\) is replaced by the metric \(d(x,y)\).
To prove part (ii), let \(T\) be an infinite subset of \(S\). Assume that no points of \(S\) can be an accumulation point of \(T\). We are going to lead to a contradiction. For each \(x\in S\), there exists an open ball \(B_{M}(x)\) satisfying
\begin{align}
& B_{M}(x;r)\cap T\nonumber\\ & \quad =\left\{\begin{array}{ll}
\emptyset & \mbox{if \(x\not\in T\)}\\ \{x\} & \mbox{if \(x\in T\)}.
\end{array}\right .{\label{ma83}}\tag{7}
\end{align}
Then, the union \(\bigcup_{x\in S}B(x)\) is an open covering of \(S\). The compactness says that there exists a finite subcollection \(\bigcup_{k=1}^{p}B(x_{k})\) that covers \(S\), which also covers \(T\). This is a contradiction by referring to (\ref{ma83}), since \(T\) is an infinite set. This completes the proof. \(\blacksquare\)
When \(M=\mathbb{R}^{n}\), the two properties in Theorem \ref{mat428} are equivalent by Theorem A in page point set topology in \(\mathbb{R}^{n}\). However, in the general metric space, property (i) cannot be equivalent to the compactness. However, we can show that property (ii) is equivalent to the compactness.
Proposition. \(S\) is a compact subset of a metric space \((M,d)\) if and only if every infinite subset of \(S\) has an accumulation point of \(S\).
Proof. The proof is left as an exercise. \(\blacksquare\)
Theorem A. Let \(S\) be a closed subset of a metric space \((M,d)\). Suppose that \((M,d)\) is itself compact. Then \(S\) is also compact.
Proof. Let \({\cal F}\) be an open covering of \(S\). Since \(S\) is closed in \(M\), we see that \(S^{c}\) is open in \(M\). This also says that \({\cal F}\cup{S^{c}}\) is an open covering of \(M\). Since \(M\) is compact, there is a finite subcovering covers \(M\). We can assume that this finite subcovering includes \(S^{c}\). In other words, we have
\begin{align*} S & \subseteq M\\ & \subseteq A_{1}\cup\cdots\cup A_{p}\cup S^{c}.\end{align*}
Therefore, we conclude that \(S\subseteq A_{1}\cup\cdots\cup A_{p}\), and the proof is complete. \(\blacksquare\)
Definition. Let \(S\) be a subset of a metric space \((M,d)\). A point \(x\in M\) is called a boundary point of \(S\) when every open ball \(B_{M}(a;r)\) contains points of \(S\) and points
of \(S^{c}\), i.e.,
\[B_{M}(a;r)\cap S\neq\emptyset\]
and
\[B_{M}(a;r)\cap S^{c}\neq\emptyset.\]
The set of all boundary points of \(S\) is called the boundary of \(S\), and is denoted by \(\partial S\). \(\sharp\)
Proposition. Let \(S\) be a subset of a metric space \((M,d)\). Then
\[\partial S=\mbox{cl}(S)\cap\mbox{cl}(S^{c}),\]
which also says that \(\partial S\) is a closed set in \(M\).
Proof. The proof is left as an exercise. \(\blacksquare\)
Example. We provide some interesting examples.
(i) Let \(M=\mathbb{R}^{n}\). We define
\[d({\bf x},{\bf y})=\parallel {\bf x}-{\bf y}\parallel\]
for any \({\bf x},{\bf y}\in\mathbb{R}^{n}\). Then, the boundary point of open ball \(B_{\mathbb{R}}({\bf a};r)\) is the set of points satisfying \(\parallel {\bf x}-{\bf a}\parallel =r\), i.e.,
\[\partial B_{\mathbb{R}}({\bf a};r)=\left\{{\bf x}\in\mathbb{R}^{n}:
\parallel {\bf x}-{\bf a}\parallel =r\right\}.\]
(ii) Let \(M=\mathbb{R}\) be the real line. We have \(\partial\mathbb{R}=\mathbb{R}\), since \(\mbox{cl}(\mathbb{Q})=\mathbb{R}\) and \(\mbox{cl}(\mathbb{Q}^{c})=\mathbb{R}\). In other words, the boundary of the set of all rational numbers in \(\mathbb{R}\) is all of points in \(\mathbb{R}\). \(\sharp\)
Connectedness.
We are going to study the connectedness of sets and its relation to continuity.
Definition. A metric space \((M,d)\) is called {\bf disconnected} when \(M=A\cup B\) such that \(A\) and \(B\) are disjoint nonempty open sets in \(M\). We call \((M,d)\) connected when it is not disconnected. A subset \(X\) of \(M\) is called connected when it is regarded as a connected metric subspace. \(\sharp\)
Example. We have the following interesting examples.
- The metric space \(M=\mathbb{R}\setminus\{0\}\) with the usual Euclidean metric is disconnected, since it is the union of two disjoint nonempty open sets, the set of all positive real numbers and the set of all negative real numbers.
- Every open interval in \(\mathbb{R}\) is connected.
- The set \(\mathbb{Q}\) of all rational numbers is disconnected. In fact, we have \(\mathbb{Q}=A\cup B\), where \(A\) consists of all rational numbers \(<\sqrt{2}\) and \(B\) consists of all rational numbers \(>\sqrt{2}\).
- Every metric space \((M,d)\) contains nonempty connected subsets. In fact, for each \(p\in M\), the singleton set \(\{p\}\) is connected. \(\sharp\)
Definition. A real-valued function \(f\) that is continuous on a metric space \((M,d)\) is said to be two-valued on \(M\) when \(f(M)\subseteq\{0,1\}\). \(\sharp\)
Example. We consider the discrete metric space \(\hat{M}=\{0,1\}\), where \(\hat{M}\) has a discrete metric. We recall that every subset of a discrete metric space \(\hat{M}\) is both open and closed in \(\hat{M}\). Therefore, the two-valued function \(f:M\rightarrow\hat{M}\) is continuous. \(\sharp\)
\begin{equation}{\label{mat443}}\tag{8}\mbox{}\end{equation}
Theorem \ref{mat443}. A metric space \((M,d)\) is connected if and only if the two-valued function on \(M\) is constant.
Proof. Assume that \(M\) is a connected set. Let \(f\) be a two-valued function defined on \(M\). We want to show that \(f\) is constant. Let \(A=f^{-1}(\{0\})\) and \(B=f^{-1}(\{1\})\). Since \(\{0,1\}\) is a discrete metric space, it follows that the singleton sets \(\{0\}\) and \(\{1\}\) are open. Since \(f\) is continuous, using Theorem \ref{mat437}, we see that \(A\) and \(B\) are open in \(M\) satisfying \(M=A\cup B\), where \(A\) and \(B\) are disjoint. Since \(M\) is connected, we conclude that either \(A=\emptyset\) and \(B=M\), or \(B=\emptyset\) and \(A=M\). In their case, we conclude that \(f\) is constant on \(M\). Conversely, assume that \(M\) is disconnected, i.e., \(M=A\cup B\), where \(A\) and \(B\) are disjoint nonempty open subsets of \(M\). We shall prove that a two-valued function on \(M\) is not constant. Let
\[f(x)=\left\{\begin{array}{ll}
0 & \mbox{if \(x\in A\)},\\
1 & \mbox{if \(x\in B\)}.
\end{array}\right .\]
Since \(A\) and \(B\) are nonempty, the function \(f\) takes values \(0\) and \(1\), which says that \(f\) is not constant. We remain to show that \(f\) is continuous. Since \(\{0,1\}\) is a discrete metric space, the inverse image of every open subset of \(\{0,1\}\) under \(f\) is open. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map444}}\tag{9}\mbox{}\end{equation}
Proposition \ref{map444}. We consider the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2})\). Let \(X\) be a connected subset of \(M_{1}\). Suppose that \(f\) is continuous on \(X\). Then \(f(X)\) is a connected subset of \(M_{2}\).
Proof. Let \(g\) be a two-valued function defined on \(f(X)\). According to Theorem \ref{mat443}, we want to show that \(g\) is constant. We consider the composite function \(h\) defined on \(X\) by \(h(x)=g(f(x))\). Then \(h\) is continuous on \(X\) and can only take the values \(0\) and \(1\), which says that \(h\) is a two-valued function on \(X\). Since \(X\) is connected, the function \(h\) is constant on \(X\), which implies that \(g\) is constant on \(f(X)\). Therefore, the set \(f(X)\) is connected. This completes the proof. \(\blacksquare\)
Example. Since an interval \(I\) in \(\mathbb{R}\) is connected, every continuous image \(f(I)\) is connected. Let \(f\) be a real-valued function. Then, the image \(f(I)\) is also an interval. \(\sharp\)
Theorem. (Intermediate-Value Theorem). Let \(f\) be a continuous real-valued function defined on a connected subset \(S\) of \(\mathbb{R}^{n}\). Suppose that \(f\) takes on two different values \(a\) and \(b\) in \(S\). Then, for each real number \(c\) between \(a\) and \(b\), there exists a point \({\bf x}\in S\) satisfying \(f({\bf x})=c\).
Proof. Using Proposition \ref{map444}, the image \(f(S)\) is a connected subset of \(\mathbb{R}\), which also says that \(f(S)\) is an interval containing \(a\) and \(b\). Therefore, if there exists \(c\) that is between \(a\) and \(b\) and is not in \(f(S)\), then \(f(S)\) would be disconnected. This contradiction completes the proof. \(\blacjsquare\)
\begin{equation}{\label{map445}}\tag{10}\mbox{}\end{equation}
Proposition \ref{map445}. Let \({\cal F}\) be a collection of connected subsets of a metric space \((M,d)\) such that the intersection \(T=\bigcap_{A\in {\cal F}}A\) is nonempty. Then, the union \(U=\bigcup_{A\in {\cal F}}A\) is connected.
Proof. Since \(T\neq\emptyset\), we take \(t\in T\). Let \(f\) be a two-valued function defined on \(U\). We are going to show that \(f\) is constant on \(U\) by claiming \(f(x)=f(t)\) for all \(x\in U\). For \(x\in U\), we have \(x\in A\) for some \(A\in {\cal F}\). Since \(A\) is connected, the function \(f\) is constant on \(A\). Since \(t\in A\), we obtain \(f(x)=f(t)\). This completes the proof. \(\blacksquare\)
Every point \(x\) in a metric space \((M,d)\) belongs to at least one connected subset of \(M\), i.e., the singleton set \(\{x\}\). By Proposition \ref{map445}, the union of all the connected subsets that contain \(x\) is also connected. We call this union a {\bf component} of \(M\), and it is denoted by \(U(x)\). In other words, \(U(x)\) is the maximal connected subset of \(S\) which contains \(x\).
\begin{equation}{\label{ma136}}\tag{11}\mbox{}\end{equation}
Proposition \ref{ma136}. Every point of a metric space \((M,d)\) belongs to a uniquely determined component of \(M\). In other words, the components of \(M\) form a collection of disjoint sets whose union is \(M\).
Proof. By Proposition \ref{map445}, two distinct components cannot contain a point \(x\). Otherwise, their union would be a larger connected set containing \(x\), which contradicts the concept of component. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma134}}\tag{12}\mbox{}\end{equation}
Definition \ref{ma134}. A subset \(S\) of \(\mathbb{R}^{n}\) is called arcwise connected or pathwise connected when, given any two points \({\bf a}\) and \({\bf b}\)in \(S\), there exists a continuous function \({\bf f}:[0,1]\rightarrow S\) satisfying \({\bf f}(0)={\bf a}\) and \({\bf f}(1)={\bf b}\). \(\sharp\)
The function \({\bf f}\) is Definition \ref{ma134} is called a {\bf path} from \({\bf a}\) to \({\bf b}\). If \({\bf f}(0)\neq {\bf f}(1)\), the image of \([0,1]\) under the function \({\bf f}\) is called an arc joining \({\bf a}\) and \({\bf b}\). In other words, \(S\) is arcwise connected when every pair of distinct points in \(S\) can be joined by an arc lying in \(S\). When
\[{\bf f}(t)=t{\bf b}+(1-t){\bf a}\mbox{ for }0\leq t\leq 1,\]
the arc joining \({\bf a}\) and \({\bf b}\) is called a line segment. It is clear to see that every convex set in \(\mathbb{R}^{n}\) is arcwise connected, since the line segment joining any two points in such set lies in the set.
Proposition. Every arcwise connected set in \(\mathbb{R}^{n}\) is connected.
Proof. Let \(S\) be an arcwise connected set, and let \(g\) be two-valued function defined on \(S\). We are going to prove that \(g\) is a constant on \(S\) by referring to Theorem \ref{mat443}. Given a point \({\bf a}\in S\), for any \({\bf x}\in S\), we can join \({\bf a}\) and \({\bf x}\) by an arc \(\Gamma\) lying in \(S\) by definition. Since the arc \(\Gamma\) is connected, Theorem \ref{mat443} says that the two-valued function \(g\) is constant on \(\Gamma\), i.e., \(g({\bf x})=g({\bf a})\). Since \({\bf x}\) can be any points in \(S\), this shows that \(g\) is constant on \(S\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma135}}\tag{13}\mbox{}\end{equation}
Proposition \ref{ma135}. Every open connected set in \(\mathbb{R}^{n}\) is arcwise connected.
Proof. Let \(S\) be an open connected set in \(\mathbb{R}^{n}\). Given any \({\bf x}\in S\), let \(A\) denote a subset of \(S\) such that \({\bf x}\) can be joined to every point \({\bf y}\in A\) by an arc lying in \(S\). Let \(B=S\setminus A\). Then \(S=A\cup B\) and \(A\cap B=\emptyset\). We shall show that \(A\) and \(B\) are open in \(\mathbb{R}^{n}\). Given \({\bf a}\in A\), assume that \({\bf a}\) is joined to \({\bf x}\) by an arc \(\Gamma\) lying in \(S\). Since \({\bf a}\in S\) and \(S\) is open, there exists an open ball \(B({\bf a};\epsilon)\) satisfying \(B({\bf a};\epsilon)\subset S\). Since \(B({\bf a};\epsilon)\) is convex, every point \({\bf y}\in B({\bf a};\epsilon)\) can be joined to \({\bf a}\) by a line segment \(L\subset B({\bf a};\epsilon)\). It follows that \({\bf y}\) can be joined to \({\bf x}\) by an arc \(L\cup\Gamma\) lying in \(S\). This says that \({\bf y}\in A\), which implies \(B({\bf a};\epsilon)\subset A\). This shows that \(A\) is open. Given \({\bf b}\in B\subset S\), since \(S\) is open, there exists an open ball \(B({\bf b};\epsilon)\) satisfying \(B({\bf b};\epsilon)\subset S\). Suppose that a point \({\bf y}\in B({\bf b};\epsilon)\) can be joined to \({\bf x}\) by an arc \(\Gamma\) lying in \(S\). Since the point \({\bf b}\) can also be joined to \({\bf y}\) by a line segment \(L\subset B({\bf b};\epsilon)\), which says that \({\bf b}\) can be joined to \({\bf x}\) by an arc \(\Gamma\cup L\) lying in \(S\). This also says that \({\bf b}\in A\), i.e., \({\bf b}\in A\cap B\). Since \(A\cap B=\emptyset\), this contradiction says that each point of \(B({\bf b};\epsilon)\) cannot be in \(A\). Since \(B({\bf b};\epsilon)\subset S=A\cup B\), it follows \(B({\bf b};\epsilon)\subset B\). This shows that \(B\) is open. Now, we obtain a decomposition \(S=A\cup B\) such that \(A\) and \(B\) are disjoint open sets in \(\mathbb{R}^{n}\). Since \({\bf x}\i A\), we have \(A\neq\emptyset\). Since \(S\) is connected, it follows \(B=\emptyset\), which shows that \(S=A\) and \(A\) is arcwise connected. This completes the proof. \(\blacksquare\)
Definition. A path \({\bf f}:[0,1]\rightarrow S\) is said to be polygonal when the image of \([0,1]\) under \({\bf f}\) is the union of a finite number of line segments. \(\sharp\)
\begin{equation}{\label{ma137}}\tag{14}\mbox{}\end{equation}
Proposition \ref{ma137}. Every open connected set \(S\) in \(\mathbb{R}^{n}\) is polygonally connected. In other words, every pair of points in \(S\) can be joined by a polygonal arc lying in \(S\).
Proof. The same argument is the proof of Proposition \ref{ma135} is still valid. \(\blacksquare\)
\begin{equation}{\label{ma18}}\tag{15}\mbox{}\end{equation}
Theorem \ref{ma18}. (The Lindelof Covering Theorem) Let \(A\) be a subset of \(\mathbb{R}^{n}\), and let \({\cal F}\) be an open covering of \(A\). Then, there exists a countable subcollection of \({\cal F}\) which also covers \(A\).
The above Theorem \ref{ma18} can refer to the page Point set topology in Euclidean Space.
Theorem. Every open set in \(\mathbb{R}^{n}\) can be expressed in one and only one way as a countable disjoint union of open connected sets.
Proof. Let \(S\) be an open set in \(\mathbb{R}^{n}\). Proposition \ref{ma135} says that the components of \(S\) form a collection of disjoint sets whose union is \(S\). Given a component \(U\) of \(S\) and any \({\bf x}\in U\), since \(S\) is open, there exists an open ball \(B({\bf x};\epsilon)\) satisfying \(B({\bf x};\epsilon)\subset S\). Since \(B({\bf x};\epsilon)\) is connected and \(U\) is the maximal connected set containing \({\bf x}\), it follows \(B({\bf x};\epsilon) \subset U\). This shows that \(U\) is an open connected set. Using the Lindel\”{o}f covering Theorem \ref{ma18}, the components of \(S\) form a countable collection, where the components are open connected sets. Proposition \ref{ma135} also says that this decomposition into components is unique. This completes the proof. \(\blacksquare\)
