Salvatore Fergola (1799-1874) was an Italian painter
A point in two-dimensional space is an ordered pair of real numbers denoted by \((x_{1},x_{2})\). Similarly, a point in three-dimensional space is an ordered triple of real numbers \((x_{1},x_{2},x_{3})\). In general, a point of \(n\)-dimensional space is an ordered \(n\)-tuple or real numbers \((x_{1},x_{2},\cdots ,x_{n})\), where each \(x_{i}\in\mathbb{R}\) for \(i=1,\cdots ,n\). Usually, it is denoted by the bold-face letter \({\bf x}\) and is sometimes called as a vector. The real number \(x_{k}\) is called the \(k\)th coordinate of the point \({\bf x}\) or the \(k\)th component of the vector \({\bf x}\). The set of all \(n\)-dimensional vectors is called $n$-dimensional Euclidean space, and is denoted by \(\mathbb{R}^{n}\). When \(n=1\), we simply write \(\mathbb{R}\) to denote the one-dimensional Euclidean space, i.e., the real line or real number system. We also see that \(\mathbb{R}^{2}=\mathbb{R}\times\mathbb{R}\) and \(\mathbb{R}^{3}=\mathbb{R}^{2}\times\mathbb{R}=\mathbb{R}\times\mathbb{R}\times\mathbb{R}\). In general, we have
\[\mathbb{R}^{n}=\overbrace{\mathbb{R}\times\cdots\times\mathbb{R}}^{\scriptsize \mbox{$n$ times}}.\]
For any two vector \({\bf x}\) and \({\bf y}\) in \(\mathbb{R}^{n}\), we define some operations as follows:
- The sum is defined by \[\alpha {\bf x}=(\alpha x_{1},\cdots ,\alpha x_{n})\]for \(\alpha\in\mathbb{R}\).
- The difference is defined by \[{\bf x}-{\bf y}={\bf x}+(-{\bf y})=\left (x_{1}-y_{1},\cdots ,x_{n}-y_{n}\right ).\]
- The zero vector is written by \({\bf 0}=(0,0,\cdots ,0)\), which is also called the origin of \(\mathbb{R}^{n}\).
- The inner product or dot product is defined by \[{\bf x}\bullet {\bf y}=\sum_{k=1}^{n}x_{k}y_{k}.\]
- The Euclidean norm of \({\bf x}\) is defined by \[\parallel {\bf x}\parallel =\sqrt{x_{1}^{2}+\cdots +x_{n}^{2}}.\]
- The distance between \({\bf x}\) and \({\bf y}\) is defined by \[\parallel {\bf x}-{\bf y}\parallel=\sqrt{(x_{1}-y_{1})^{2}+\cdots+(x_{n}-y_{n})^{2}}.\]
The unit coordinate vector \({\bf e}_{k}\) in \(\mathbb{R}^{n}\) is the vector whose \(k\)th component is \(1\) and whose remaining components are zero. For example,
\[\begin{array}{llll}{\bf e}_{1}=(1,0,\cdots ,0), & {\bf e}_{2}=(0,1,0,\cdots ,0), & \cdots , & {\bf e}_{n}=(0,0,\cdots ,0,1).\end{array}\]
It is easy to see that any vector \({\bf x}=(x_{1},\cdots ,x_{n})\) can be represented as
\[{\bf x}=x_{1}{\bf e}_{1}+x{2}{\bf e}_{2}+\cdots +x_{n}{\bf e}_{n}\]
and
\[\begin{array}{llll}
x_{1}={\bf x}\bullet {\bf e}_{1}, & x_{2}={\bf x}\bullet {\bf e}_{2}, & \cdots , & x_{n}={\bf x}\bullet {\bf e}_{n}.
\end{array}\]
The vectors \({\bf e}_{1},{\bf e}_{2},\cdots ,{\bf e}_{n}\) are also called the basis vectors.
Proposition. Let \({\bf x}\) and \({\bf y}\) be any two points in \(\mathbb{R}^{n}\).
(i) We have \(\parallel {\bf x}\parallel\geq 0\). We also have \(\parallel {\bf x}\parallel =0\) if and only if \({\bf x}={\bf 0}\).
(ii) For any \(\alpha\in\mathbb{R}\), we have \(\parallel\alpha {\bf x}\parallel=|\alpha |\parallel {\bf x}\parallel\). It also says that \(\parallel {\bf x}-{\bf y}\parallel =\parallel {\bf y}-{\bf x}\parallel\).
(iii) We have the Cauchy inequality \(\left |{\bf x}\bullet {\bf y}\right |\leq\parallel {\bf x}\parallel\parallel {\bf y}\parallel\).
(iv) We have the triangle inequality \(\parallel {\bf x}+{\bf y}\parallel\leq\parallel {\bf x}\parallel +\parallel {\bf y}\parallel\).
Definition. Given any point \({\bf a}\) in \(\mathbb{R}^{n}\) and any positive real number \(r\), the \(n\)-dimensional open ball of radius \(r\) and center \({\bf a}\) is defined by
\[B({\bf a};r)=\left\{{\bf x}\in\mathbb{R}^{n}:\parallel {\bf x}-{\bf a}\parallel <r\right\}.\]
We have the following easy observations.
- In \(\mathbb{R}\), the open ball is an open interval.
- In \(\mathbb{R}^{2}\), the open ball is a circle of radius \(r\) and center \({\bf a}\) without the boundary.
- In \(\mathbb{R}^{3}\), the open ball is a spherical solid of radius \(r\) and center \({\bf a}\) without the boundary.
Definition. Let \(S\) be a subset of \(\mathbb{R}^{n}\). Given any \({\bf a}\in S\), we say that \({\bf a}\) is an interior point of \(S\) when there is an open ball \(B({\bf a};r)\) satisfying \(B({\bf a};r)\subseteq S\). The set of all interior points of \(S\) is called the interior of \(S\), and is denoted by \(\mbox{int}(S)\). We say that \(S\) is open when \(S=\mbox{int}(S)\). \(\sharp\)
It is easy to see that \(\mbox{int}(S)\subseteq S\). Let \(I=[a,b]\) be a closed interval in \(\mathbb{R}\). It is clear to see that the interior of \(I\) is an open interval \((a,b)\), i.e., \(\mbox{int}(I)=(a,b)\). We also have the following observations.
- Let \(S\) be an open subset of \(\mathbb{R}^{n}\). Then, given any \({\bf a}\in S\), there is an open ball \(B({\bf a};r)\) satisfying \(B({\bf a};r)\subseteq S\).
- Suppose that \(S\subseteq\mbox{int}(S)\). Then \(S\) is an open set, since \(\mbox{int}(S)\subseteq S\) in general.
- The open interval in \(\mathbb{R}\) is an open set in \(\mathbb{R}\).
- The whole Euclidean space \(\mathbb{R}^{n}\) is an open set.
- The empty set \(\emptyset\) is an open set.
- Each open ball is an open set.
\begin{equation}{\label{map418}}\tag{1}\mbox{}\end{equation}
Proposition \ref{map418}. The union of any collection of open sets is an open set.
Proof. Let \({\cal F}\) be any collection of open sets in \(\mathbb{R}^{n}\), and let \(S\) denote the union \(S=\bigcup_{A\in {\cal F}}A\). Given any \({\bf x}\in S\), we see that \({\bf x}\in A\) for some \(A\in {\cal F}\). Since \(A\) is open, there exists an open ball \(B({\bf x};r)\subseteq A\). Since \(A\subseteq S\), we also have \(B({\bf x};r)\subseteq S\), which shows that \({\bf x}\in
\mbox{int}(S)\), i.e., \(S\subseteq\mbox{int}(S)\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map419}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map419}. The intersection of a finite collection of open sets is open.
Proof. Let \(S=\bigcap_{k=1}^{n}A_{k}\), where each \(A_{k}\) is open for \(k=1,\cdots ,n\). If \(S=\emptyset\), then it is open. Therefore, we assume \(S\neq\emptyset\). Given any \({\bf x}\in S\), we have \({\bf x}\in A_{k}\) for each \(k=1,\cdots ,n\). Since each \(A_{k}\) is open, there exists \(r_{k}\) satisfying \(B({\bf x};r_{k})\subseteq A_{k}\) for \(k=1,\cdots ,n\). Let \(r=\min\{r_{1},\cdots ,r_{n}\}\). Then, we have \(B({\bf x};r)\subseteq B({\bf x};r_{k})\) for all \(k=1,\cdots ,n\), which says \({\bf x}\in B({\bf x};r)\subseteq S\). This shows \({\bf x}\in\mbox{int}(S)\), i.e., \(S\subseteq\mbox{int}(S)\). This completes the proof. \(\blacksquare\)
Arbitrary intersections cannot lead to an open set in general. For example, the intersection of open intervals of the form \((-1/n,1/n)\) for \(n=1,2,\cdots\) is the set consisting of 0 alone, which is not an open set.
Definition. A subset \(S\) of \(\mathbb{R}^{n}\) is called a closed set when its complement \(S^{c}=\mathbb{R}^{n}\setminus S\) is open. \(\sharp\)
A closed interval \([a,b]\) in \(\mathbb{R}\) is a closed set.
\begin{equation}{\label{map420}}\tag{3}\mbox{}\end{equation}
Proposition \ref{map420}. The union of finite collection of closed sets is a closed set, and the intersection of any collection of closed sets is closed.
Proof. Since we have
\[\left (\bigcup_{k=1}^{n}A_{k}\right )^{c}=\bigcap_{k=1}^{n}A^{c}\]
and
\[\left (\bigcap_{A\in {\cal F}}A\right )^{c}=\bigcup_{A\in {\cal F}}A^{c},\]
the results follow immediately from Propositions \ref{map418} and \ref{map419}. \(\blacksquare\)
Proposition. Suppose that \(A\) is open and \(B\) is closed. Then \(A\setminus B\) is open and \(B\setminus A\) is closed.
Proof. Since we have
\[A\setminus B=A\cap B^{c}\]
and
\[B\setminus A=B\cap A^{c},\]
the results follow immediately from Propositions \ref{map419} and \ref{map420}. \(\blacksquare\)
Definition. Let \(S\) be a subset of \(\mathbb{R}^{n}\).
- The point \({\bf x}\) in \(\mathbb{R}^{n}\) is said to be an adherent point of \(S\) when every open ball \(B({\bf x};r)\) contains at least one point of \(S\), i.e., \(B({\bf x};r)\cap S\neq\emptyset\).
- The point \({\bf x}\) in \(\mathbb{R}^{n}\) is said to be an accumulation point of \(S\) when every open ball \(B({\bf x};r)\) satisfies \(B({\bf x};r)\setminus\{{\bf x}\}\cap S\neq\emptyset\).
- If \({\bf x}\in S\) and \({\bf x}\) is not an accumulation point of \(S\), then \({\bf x}\) is called an isolated point. \(\sharp\)
We have the following observations.
- Every point \({\bf x}\) in \(S\) is an adherent point of \(S\).
- \({\bf x}\) is an accumulation point of \(S\) if and only if \({\bf x}\) is an adherent point of \(S\setminus\{{\bf x}\}\).
Example. We consider the real number system \(\mathbb{R}\).
- Let \(S\) be a subset of \(\mathbb{R}\) such that \(S\) is bounded above. Then \(\sup S\) is an adherent point of \(S\).
- The set \({1,1/2,\cdots ,1/n,\cdots}\) has 0 as an accumulation point.
- The set of rational numbers has every real number as an accumulation point. In other words, given any real number \(x\), there exists a sequence of rational numbers \(\{r_{n}\}_{n=1}^{\infty}\) satisfying \(r_{n}\rightarrow x\).
- Every point of the closed interval \([a,b]\) is an accumulation point of open interval \((a,b)\).
\begin{equation}{\label{map421}}\tag{4}\mbox{}\end{equation}
Proposition \ref{map421}. Let \({\bf x}\) be an accumulation point of \(S\). Then, every open ball \(B({\bf x};r)\) contains infinitely many points of \(S\). In particular, the finite set cannot have the accumulation points.
Proof. Suppose that \(B({\bf x};r)={{\bf a}_{1},\cdots ,{\bf a}_{n}}\) contains finite number of points, where \({\bf a}_{i}\neq {\bf x}\) for \(i=1,\cdots ,n\). We are going to lead to a contradiction. Let
\[r_{0}=\min\left\{\parallel {\bf x}-{\bf a}_{1}\parallel ,\cdots , \parallel {\bf x}-{\bf a}_{n}\parallel\right\}.\]
Then \(B({\bf x};r_{0}/2)\) is an open ball containing no points of \(S\) distinct from \({\bf x}\). This contradicts the definition of accumulation point, and the proof is complete. \(\blacksquare\)
It is possible that an infinite set cannot have an accumulation point. For example, the infinite set \(\{1,2,3,\cdots\}\) does not have an accumulation point.
\begin{equation}{\label{mat422}}\tag{5}\mbox{}\end{equation}
Theorem \ref{mat422}. The subset \(S\) of \(\mathbb{R}^{n}\) is closed if and only if \(S\) contains all its adherent points.
Proof. Suppose that \(S\) is closed. Let \({\bf x}\) be an adherent point of \(S\). We wish to show \({\bf x}\in S\). Assume \({\bf x}\not\in S\). We want to lead to a contradiction. Since \({\bf x}\in S^{c}\), and \(S^{c}\) is open, there exists an open ball \(B({\bf x};r)\) satisfying \(B({\bf x};r)\subset S^{c}\). This also says \(B({\bf x};r)\cap S=\emptyset\), which contradicts \(B({\bf x};r)\cap S\neq\emptyset\).
For the converse, we assume that \(S\) contains all its adherent points. For \({\bf x}\in S^{c}\), it means that \({\bf x}\) is not an adherent point of \(S\). By definition, there exists an
open ball \(B({\bf x};r)\) satisfying \(B({\bf x};r)\cap S=\emptyset\), which implies \(B({\bf x};r)\subset S^{c}\). This shows that \(S^{c}\) is open, i.e., \(S\) is closed. This completes the proof. \(\blacksquare\)
Definition. Let \(S\) be a subset of \(\mathbb{R}^{n}\).
- 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 of \(S\), and is denoted by \(S’\). \(\sharp\)
We have the following observations.
- Since each point of \(S\) is an adherent point of \(S\), we have \(S\subseteq\mbox{cl}(S)\).
- We have \(\mbox{cl}(S)=S\cup S’\).
\begin{equation}{\label{ma78}}\tag{6}\mbox{}\end{equation}
Proposition. Let \(S\) be a subset of \(\mathbb{R}^{n}\).
(i) \(S\) is closed if and only if \(S=\mbox{cl}(S)\).
(ii) \(S\) is closed if and only if \(S’\subseteq S\). In other words, \(S\) is closed if and only if \(S\) contains all its accumulation points.
Proof. Theorem \ref{mat422} says \(\mbox{cl}(S)\subseteq S\). Since \(S\subseteq\mbox{cl}(S)\), we obtain \(S=\mbox{cl}(S)\). This proves part (i). Part (ii) follows from part (i) and
the fact of \(\mbox{cl}(S)=S\cup S’\). This completes the proof.
Definition. A subset \(S\) of \(\mathbb{R}^{n}\) is said to be bounded when there exists an open ball \(B({\bf x};r)\) satisfying \(S\subseteq B({\bf x};r)\) for some \(r>0\) and some \({\bf x}\in\mathbb{R}^{n}\). \(\sharp\)
\begin{equation}{\label{mat431}}\tag{7}\mbox{}\end{equation}
Theorem. \ref{mat431} (The Bolzano-Weierstrass Theorem) Suppose that a bounded subset \(S\) of \(\mathbb{R}^{n}\) contains infinitely many points. Then, there exists at least one accumulation point of \(S\).
\begin{equation}{\label{ma19}}\tag{8}\mbox{}\end{equation}
Theorem \ref{ma19}. (Cantor Intersection Theorem) Let \({Q_{1},Q_{2},\cdots}\) be a countable collection of nonempty subsets of \(\mathbb{R}^{n}\) such that the following conditions are satisfied:
- \(Q_{k+1}\subseteq Q_{k}\) for \(k=1,2,\cdots\);
- \(Q_{1}\) is bounded and each set \(Q_{k}\) is closed for \(k=1,2,\cdots\).
Then, the intersection \(\bigcap_{k=1}^{\infty}Q_{k}\) is closed and nonempty.
Proof. It clear to see that \(\bigcap_{k=1}^{\infty}Q_{k}\) is a closed set by Proposition \ref{map420}. Since \(Q_{k+1}\subseteq Q_{k}\) for \(k=1,2,\cdots\), if \(Q_{k}\) consists of finitely many points for some \(k\), then the proof is obvious. Therefore, we assume that each \(Q_{k}\) consists of infinitely many points. Since \(Q_{k+1}\subseteq Q_{k}\), we can also form a collection of distinct points \(A={{\bf x}_{1},{\bf x}_{2},\cdots}\) satisfying \({\bf x}_{k}\in Q{k}\). Then, \(A\) consists of infinitely many points. Since \(A\) is contained in a bounded set \(Q_{1}\), the Bolzano-Weierstrass Theorem \ref{mat431} says that \(A\) has an accumulation point that is assumed to be a point \({\bf x}\). From Proposition \ref{map421}, every open ball of \({\bf x}\) contains infinitely many points of \(A\). Since \(Q_{k+1}\subseteq Q_{k}\), i.e., \({{\bf x}_{k},{\bf x}_{k+1},\cdots}\subset Q_{k}\), it follows that this open ball of \({\bf x}\) also contains points of \(Q_{k}\), which says that \({\bf x}\) is an accumulation point of \(Q_{k}\) for all \(k=1,2,\cdots\). Since each \(Q_{k}\) is closed, we obtain \({\bf x}\in Q_{k}\) by Proposition \ref{ma78} for all \(k=1,2,\cdots\). This completes the proof. \(\blacksquare\)
Definition. Let \(S\) be a subset of \(\mathbb{R}^{n}\). A collection \({\cal F}\) of sets is said to be a covering of \(S\) when \(S\subseteq\bigcup_{A\in {\cal F}}A\). If \({\cal F}\) is a collection
of open sets, then \({\cal F}\) is called an open covering of \(S\). \(\sharp\)
Example. We have the following examples.
(i) The collection of all intervals of the form \((1/n,2/n)\) for \(n=2,3,\cdots\) is an open covering of the open interval \((0,1)\). This is also a countable covering.
(ii) The real line \(\mathbb{R}\) is covered by the collection of all open intervals \((a,b)\). This is an uncountable covering. However, all intervals of the form \((n,n+2)\) for \(n\in\mathbb{Z}\) is a countable covering of \(\mathbb{R}\).
(iii) Let \(S=\{(x,y):x>0,y>0\}\subseteq\mathbb{R}^{2}\). The collection of all \(2\)-dimensional open ball \(B((x,x);x)\) with centers \((x,x)\) and radius \(x\) for \(x>0\) is a covering of \(S\). This covering is uncountable. If we take the collection of \(2\)-dimensional open ball \(B((x,x);x)\) satisfying \(x\in\mathbb{Q}\), then this is a countable covering. \(\sharp\)
\begin{equation}{\label{ma17}}\tag{9}\mbox{}\end{equation}
Lemma \ref{ma17}. We consider the following countable collection of open balls
\[{\cal G}=\left\{B({\bf x};r):x_{i},r\in\mathbb{Q}\mbox{ for }i=1,\cdots ,n\right\}.\]
Given \({\bf z}\in\mathbb{R}^{n}\), let \(S\) be an open subset of \(\mathbb{R}^{n}\) satisfying \({\bf z}\in S\). Then, there exists an open ball in \({\cal G}\) such that it contains \({\bf z}\) and is contained in \(S\). In other words, we have \({\bf z}\in B({\bf y};q)\subset S\) for some \(B({\bf y};q)\in {\cal G}\).
Proof. It is obvious that the collection \({\cal G}\) is countable. Since \({\bf z}\in S\) and \(S\) is an open set, there exists an \(n\)-dimensional open ball \(B({\bf z};r)\) satisfying \(B({\bf z};r)\subset S\). For \({\bf z}=(z_{1},\cdots ,z_{n})\), using the denseness of \(\mathbb{Q}\), there exists \(y_{k}\in\mathbb{Q}\) satisfying
\[\left |y_{k}-z_{k}\right |<\frac{r}{4n}\mbox{ for each }k=1,\cdots ,n.\]
Then, we have
\begin{align*} & \parallel {\bf y}-{\bf z}\parallel\\ & \quad\leq\left |y_{1}-z_{1}\right |+\left |y_{2}-z_{2}\right |
+\cdots +\left |y_{n}-z_{n}\right |\\ & \quad<\frac{r}{4}.\end{align*}
We can also take \(q\in\mathbb{Q}\) satisfying
\[\frac{r}{4}<q<\frac{r}{2}.\]
Then, we have \({\bf z}\in B({\bf y};q)\). Now, given any \({\bf x}\in B({\bf y};q)\), i.e., \(\parallel {\bf x}-{\bf y}\parallel<q\), we also have
\begin{align*} & \parallel {\bf x}-{\bf z}\parallel\\ & \quad\leq\parallel {\bf x}-{\bf y}\parallel
+\parallel {\bf y}-{\bf z}\parallel\\ & \quad<q+\frac{r}{4}\\ & \quad\leq\frac{r}{2}+\frac{r}{4}<r,\end{align*}
which says that \({\bf x}\in B({\bf z};r)\). Therefore, we obtain
\[B({\bf y};q)\subset B({\bf z};r)\subset S.\]
Since \(B({\bf y};q)\in {\cal G}\), this completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma18}}\tag{10}\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 \(A\).
Proof. Let \({\cal G}\) denote the countable collection of all open balls having rational centers and rational radii as given in Lemma \ref{ma17}. Given any \({\bf x}\in A\), since \({\cal F}\) is an open covering of \(A\), there is an open set \(S_{\bf x}\in {\cal F}\) satisfying \(x\in S_{\bf x}\). Lemma \ref{ma17} says that there is an open ball \(B_{\bf x}\) in \({\cal G}\) satisfying \({\bf x}\in B_{\bf x}\subset S_{\bf x}\). When \({\bf x}\) varies over all elements of \(A\), we obtain a countable collection of open balls which covers \(A\), i.e., \(A\subseteq\bigcup_{{\bf x}\in A}B_{\bf x}\) which is a countable union. Since each \(B_{x}\) corresponds to \(S_{\bf x}\in {\cal F}\) satisfying \({\bf x}\in B_{\bf x}\subset S_{\bf x}\), it follows \(A\subseteq\bigcup_{{\bf x}\in A}S_{\bf x}\), which is also a countable union. This completes the proof. \(\blacksquare\)
Definition. A subset \(S\) of \(\mathbb{R}^{n}\) is said to be compact when every open covering of \(S\) contains a finite subcovering of \(S\), i.e., there is a finite subcollection which covers \(S\). \(\sharp\)
Theorem. (The Heine-Borel Theorem) Every closed and bounded subset of \(\mathbb{R}^{n}\) is compact.
Proof. Let \({\cal F}\) be an open covering of a closed and bounded subset \(A\) of \(\mathbb{R}^{n}\). We are going to show that there is a finite subcollection of \({\cal F}\) which covers \(A\). From the Lindel\”{o}f Covering Theorem \ref{ma18}, there is a countable subcollection \({I_{1},I_{2},\cdots}\) covers \(A\). For \(m\in\mathbb{N}\), we consider the finite union
\[S_{m}=\bigcup_{k=1}^{m}I_{k}.\]
It is clear to see that \(S_{m}\) is an open set and \(A\subseteq\bigcup_{m=1}^{\infty}S_{m}\). We define a countable collection of sets \({Q_{1},Q_{2},\cdots}\) by
\[Q_{m}=A\cap S_{m}^{c}\mbox{ for }m\geq 1,\]
where \(S_{m}^{c}=\mathbb{R}^{n}\setminus S_{m}\) is a closed set. Since \(A\) is closed and bounded, it follows that \(Q_{1}\) is bounded and each \(Q_{m}\) is closed for \(m\geq 1\).
Since \(S_{m}\subseteq S_{m+1}\), we see that \(Q_{m+1}\subseteq Q_{m}\). Suppose that each \(Q_{m}\) is nonempty. We are going to lead to a contradiction. Using the Cantor Intersection Theorem \ref{ma19}, it follows
\begin{align*} & A\cap\left (\bigcup_{m=1}^{\infty}S_{m}\right )^{c}\\
& \quad =A\cap\left (\bigcap_{m=1}^{\infty}S_{m}^{c}\right )\\
& \quad =\bigcap_{m=1}^{\infty}\left (A\cap S_{m}^{c}\right )\\
& \quad =\bigcap_{m=1}^{\infty}Q_{m}\neq\emptyset,\end{align*}
which contradicts \(A\subseteq\bigcup_{m=1}^{\infty}S_{m}\). Therefore, there exists \(Q_{m_{0}}=\emptyset\), i.e.,
\[\emptyset =Q_{m_{0}}=A\cap S_{m_{0}}^{c}.\]
Then, we obtain
\[A\subseteq S_{m_{0}}=\bigcup_{k=1}^{m_{0}}I_{k}.\]
This completes the proof. \(\blacksquare\)
Theorem A. Let \(S\) be a subset of \(\mathbb{R}^{n}\). Then, the following statements are equivalent.
(a) \(S\) is compact.
(b) \(S\) is closed and bounded.
(c) Every infinite subset of \(S\) has an accumulation point of \(S\).
Proof. We are going to prove that statement (a) implies statement (b). Suppose that \(S\) is compact. Given a point \({\bf x}\in S\), the collection of open balls \(\{B({\bf x};k)\}_{k=1}^{\infty}\) is an open covering of \(S\). The compactness says that a finite subcollection \(\{B({\bf x};k)\}_{k=1}^{n}\) is also a covering of \(S\), which shows that \(S\) is bounded. Now, we assume that \(S\) is not closed. Then, there exists an accumulation point \({\bf y}\) of \(S\) satisfying \({\bf y}\not\in S\). For each \({\bf x}\in S\), let
\begin{equation}{\label{ma82}}\tag{11}
r_{\bf x}=\frac{1}{2}\parallel {\bf x}-{\bf y}\parallel.
\end{equation}
Then \(r_{\bf x}\neq 0\), since \({\bf y}\not\in S\). We also see that the collection \(\{B({\bf x};r_{\bf x}):{\bf x}\in S\}\) is an open covering of \(S\). The compactness says that there is a finite subcollection that cover \(S\). In other words, we have
\begin{equation}{\label{ma79}}\tag{12}
S\subseteq\bigcup_{k=1}^{n}B({\bf x}_{k};r_{k}),
\end{equation}
where \(r_{k}\equiv r_{{\bf x}_{k}}\). Let \(r=\min\{r_{1},\cdots,r_{n}\}\). Given any \({\bf x}\in B({\bf y};r)\), we have
\[\parallel {\bf x}-{\bf y}\parallel<r\leq r_{k}\mbox{ for all }k.\]
Using the triangle inequality, we also have
\[\parallel {\bf y}-{\bf x}_{k}\parallel\leq\parallel{\bf y}-{\bf x}\parallel +\parallel {\bf x}-{\bf x}_{k}\parallel,\]
which implies, by using (\ref{ma82}),
\begin{align*} & \parallel {\bf x}-{\bf x}_{k}\parallel\\ & \quad\geq\parallel{\bf y}-{\bf x}_{k}\parallel
-\parallel {\bf x}-{\bf y}\parallel\\ & \quad = 2r_{k}-\parallel {\bf x}-{\bf y}\parallel>r_{k}.\end{align*}
This shows \({\bf x}\not\in B({\bf x}_{k};r_{k})\) for all \(k\). Using (\ref{ma79}), it follows \(B({\bf y};r)\cap S=\emptyset\), which contradicts the fact that \({\bf y}\) is an accumulation point of \(S\). This contradiction shows that \(S\) is a closed set. Using the Heine-Borel Theorem, we see that statements (a) and (b) are equivalent.
We are going to prove that statement (b) implies statement (c). Suppose that \(S\) is closed and bounded. Let \(T\) be any infinite subset of \(S\). Then \(T\) is also bounded. Using the Bolzano-Weierstrass Theorem \ref{mat431}, \(T\) has an accumulation point \({\bf x}\), which also says that \({\bf x}\) is an accumulation point of \(S\). Since \(S\) is closed, it follows \({\bf x}\in S\).
We are going to prove that statement (c) implies statement (b). Assume that \(S\) is unbounded. Then, for each \(k>0\), there exists a point \({\bf x}_{k}\in S\) satisfying \(\parallel {\bf x}_{m}\parallel>m\). The collection \(T=\{{\bf x}_{k}\}{k=1}^{\infty}\) is an infinite subset of \(S\). Therefore, \(T\) has an accumulation point \({\bf y}\in S\) by the hypothesis. For \(k>1+\parallel{\bf y}\parallel\), we have
\begin{align*} & \parallel {\bf x}_{k}-{\bf y}\parallel\\ & \quad\geq\parallel {\bf x}_{k}\parallel-\parallel
{\bf y}\parallel\\ & \quad >m-\parallel{\bf y}\parallel>1,\end{align*}
which contradicts the fact that \({\bf y}\) is an accumulation point of \(T\). This contradiction says that \(S\) is bounded. We remain to show that \(S\) is closed. Let \({\bf x}\) be an accumulation point of \(S\). Then \(B({\bf x};1/k)\cap S\neq\emptyset\) for all \(k\). We can form an infinite set \(T={{\bf x}_{k}}{k=1}^{\infty}\) satisfying \({\bf x}_{k}\in B({\bf x};1/k)\) and \(T\subset S\). The hypothesis says that \(T\) has an accumulation point \({\bf y}\in S\). We are going to claim that \({\bf y}={\bf x}\). Assume that \({\bf y}\neq {\bf x}\). The triangle inequality says
\begin{align}
& \parallel {\bf y}-{\bf x}\parallel\nonumber\\ & \quad\leq\parallel {\bf y}-{\bf x}_{k}\parallel+\parallel {\bf x}_{k}-{\bf x}\parallel\nonumber\\ & \quad <\parallel {\bf y}-{\bf x}_{k}\parallel+\frac{1}{k}\mbox{ for } {\bf x}_{k}\in T.\label{ma80}\tag{13}\end{align}
We take a sufficiently large \(k^{*}\) satisfying
\[\frac{1}{2}\parallel {\bf y}-{\bf x}\parallel>\frac{1}{k}\mbox{ for }k\geq k^{*}.\]
Using (\ref{ma80}), we obtain
\begin{equation}{\label{ma81}}\tag{14}
\frac{1}{2}\parallel {\bf y}-{\bf x}\parallel<\parallel {\bf y}-{\bf x}_{k}\parallel.
\end{equation}
Let \(r=\frac{1}{2}\parallel {\bf y}-{\bf x}\parallel\). Using (\ref{ma81}), We see have \({\bf x}_{k}\not\in B({\bf y};r)\) for \(k\geq k^{*}\), which says that \({\bf y}\) cannot be an accumulation point of \(T\). This contradiction says \({\bf y}={\bf x}\), i.e., \({\bf x}\in S\). This completes the proof. \(\blacksquare\)
