Definition 4.16.
A cluster point of a subset \(A\) in a metric space \(X\) is an element \(x\in X\) such that there is a sequence \((x_n)_{n=1}^\infty\subseteq A\setminus \{x\}\) with \(\lim_{n\rightarrow\infty}x_n=x\text{.}\)
Section 4.2 Open and Closed Sets
Subsection Closed Sets
“I can get money with my eyes closed.”―Drake
Remark 4.17.
\(x\) need not be in \(A\)
Convention 4.18.
Cluster points are often called limit points but the book uses the term limit point slightly differently allowing \((x_n)_{n=1}^\infty\subseteq A\) instead of \((x_n)_{n=1}^\infty\subseteq A\setminus \{x\}\text{.}\)
Example 4.19.
Fix \(a,b\in \mathbb{R}, a<b\text{.}\) Set \([a,b]=\{{x\in\mathbb{R}}\,:\,{a\leq x\leq b}\}\text{.}\) The set of cluster points of \([a,b]\) is \([a,b]\text{.}\)
Solution.
Suppose first \(x\in [a,b]\text{.}\) We’ll show \(x\) is a cluster point. Suppose \(x\neq b\text{.}\) Fix an integer \(N\geq 1\) so that \(\frac{1}{N}<b-x\text{.}\) Then \((x+\frac{1}{n})_{n=N}^\infty\subseteq [a,b]\setminus\{x\}\) and \(x=\lim_{n\rightarrow\infty}(x+\frac{1}{n})\text{,}\) so \(x\) is a cluster point. For \(x=b\text{,}\) consider \((x-\frac{1}{n})_{n=M}^\infty\subseteq [a,b]\) where \(\frac{1}{M}<b-a\text{.}\)
Now suppose \(x\in \mathbb{R}\) is a cluster point of \([a,b]\text{.}\) We’ll show \(x\in [a,b]\text{.}\) We know \(\exists (x_n)_{n=1}^\infty\subseteq [a,b]\setminus\{x\}\) such that \(\lim_{n\rightarrow\infty}x_n=x\text{.}\) Since \(a\leq x_n\leq b\) for all \(n\geq 1\text{,}\) we have \(a\leq x\leq b\text{.}\) So \(x\in[a,b]\text{.}\)
Example 4.20.
For \(a,b\in \mathbb{R}\) with \(a<b\text{,}\) \([a,b]\) is the set of cluster points of
- \((a,b)=\{{x\in\mathbb{R}}\,:\,{a< x< b}\}\) (basically same proof)
- \(\displaystyle (a,b]=\{{x\in\mathbb{R}}\,:\,{a< x\leq b}\}\)
- \(\displaystyle [a,b)=\{{x\in\mathbb{R}}\,:\,{a\leq x< b}\}\)
Example 4.21.
Let \(X\) be a metric space and let \(F\subseteq X\) be a finite set. Then \(F\) has no cluster points.
Solution.
Suppose \(x\in X\) is a cluster point of \(F\text{.}\) Then \(\exists (x_n)_{n=1}^\infty\subseteq F\setminus \{x\}\) such that \(\lim_{n\rightarrow\infty} x_n=x\text{.}\) Note that \((x_n)_{n=1}^\infty\) converges so it’s Cauchy. Let \(\varepsilon=\min\{{d(y,z)}\,:\,{y,z\in F, y\neq z}\}>0\text{.}\) Fix \(N\in\mathbb{Z},N\geq 1\) such that for all \(k,l\geq N\text{,}\) \(d(x_k,x_l)<\varepsilon\text{.}\) But then \(x_k=x_l\) for all \(k,l\geq N\) and in particular \(x_k=x_N\) for all \(k\geq N\text{.}\) Then \(x=\lim_{n\rightarrow\infty}x_n=x_N\in F\setminus\{x\}\text{,}\) contradiction.
Exploration 4.22.
Almost the same proof as (3) shows that \(\mathbb{Z}\subseteq \mathbb{R}\) has no cluster points and if \(X\) is a discrete metric space then \(X\) has no cluster points.
Definition 4.23. Closed Set.
If \(X\) is a metric space and \(A\subseteq X\) is a subset, then \(A\) is closed if every cluster point of \(A\) belongs to \(A\text{.}\)
Theorem 4.24.
Let \(X\) and \(Y\) be closed subsets of a metric space. Then \(X\cup Y\) is closed.
Corollary 4.25.
If \(X_1, X_2,..., X_n\) are closed subsets of a metric space, then \(X_1\cup X_2\cup\dots\cup X_n\) is closed.
Example 4.26. Infinite Unions Need Not Be Closed.
Corollary 4.27.
Any finite subset of a metric space is closed.
Theorem 4.28.
Let \(\cC\) be a collection of closed subsets of a metric space. Then \(\bigcap\cC\) is closed.
Subsection Open Sets
“Let yourself be open and life will be easier.”―Buddha
Definition 4.29.
Let \(X\) be a metric space. The open ball of radius \(r>0\) with center \(a\in X\) is \(B_r(a)=\{{x\in X}\,:\,{d(a,x)<r}\}\text{.}\)
Example 4.30.
The open ball in \(R\) of radius ε centered at x is the open interval (x − ε, x + ε). An open ball in R2 is the interior of a disk and an open ball in R3 is the interior of a sphere.
Warning 4.31.
In general, \(\overline{B_r(a)}\neq \{{x\in X}\,:\,{d(a,x)\leq r}\}\) (they are equal in \(\mathbb{R}^n\))
Example 4.32.
\(X=\mathbb{Z}, d(x,y)=\left|{x-y}\right|\text{.}\) \(B_1(0)=\{0\}\) is closed, \(\overline{B_1(0)}=\{0\}\text{.}\) But \(\{{x\in \mathbb{Z}}\,:\,{d(x,0)\leq 1}\}=\{-1,0,1\}\)
Definition 4.33.
Let M be a metric space and let X be a subset of M. We say that X is open if for every x\in X, there exists an open ball Bε(x) (centered at x) such that B_\e(x)\sse X.
Example 4.34.
- An open interval (a, b) in R is open in R
- \(A=(0,1]\subseteq \mathbb{R}\text{,}\) \(A\) is not open since \(1\in A\) but \(B_\varepsilon(1)=(1-\varepsilon,1+\varepsilon)\) is not contained in \(A\) for any \(\varepsilon>0\text{.}\) Also \(A\) is not closed since \(0\) is a cluster point of \(A\) but \(0\notin A\text{.}\)
Theorem 4.35.
Let M be a metric space. Then M and \es are open subsets of M.
Theorem 4.36.
Open balls are open. That is, if \(a\in X\) and \(r>0\text{,}\) then \(B_r(a)\) is open.
Proof.
Fix \(x\in B_r(a)\text{.}\) Then \(d(x,a)<r\text{.}\) Let \(\varepsilon=r-d(x,a)>0\text{.}\) It's enough to show that \(B_{\varepsilon}(x)\subseteq B_r(a)\text{.}\) Fix \(y\in B_\varepsilon(x)\text{.}\) Then \(d(x,y)<\varepsilon\) and so \(d(y,a)\leq d(y,x)+d(x,a)<\varepsilon+(r-\varepsilon)=r\text{.}\) So \(y\in B_r(a)\) and \(B_{\varepsilon}(x)\subseteq B_r(a)\text{.}\)
Theorem 4.37. Properties of Open Sets.
If \(X\) is a metric space
- \(\emptyset\) and \(X\) are open
- If \(U,V\subseteq X\) are open, then \(U\cap V\) is open.
- If \((U_i)_{i\in I}\) is a collection of open subsets of \(X\text{,}\) then \(\cup_{i\in I}U_i\) is open.
Proof.
- 1 is clear (already talked through)
- Fix \(x\in U\cap V\text{.}\) Since \(U\) and \(V\) are open, there are \(\varepsilon_1,\varepsilon_2\) such that \(B_{\varepsilon_1}(x)\subseteq U\) and \(B_{\varepsilon_2}(x)\subseteq V\text{.}\) Let \(\varepsilon=\min\{\varepsilon_1,\varepsilon_2\}\text{.}\) Then \(B_{\varepsilon}(x)\subseteq B_{\varepsilon_1}(x)\subseteq U\) and \(B_{\varepsilon}(x)\subseteq B_{\varepsilon_2}(x)\subseteq V\text{.}\) So \(B_{\varepsilon}(x)\subseteq U\cap V\text{.}\)
- If \(x\in\cup_{i\in I}U_i\text{,}\) then \(x\in U_j\) for some \(j\in I\text{.}\) As \(u_j\) is open, there is an \(\varepsilon>0\) such that \(B_\varepsilon(x)\subseteq U_j\text{.}\) But then \(B_\varepsilon(x)\subseteq \cup_{i\in I}U_i\text{.}\)
Example 4.38. Arbitrary Intersection Need Not Be Open.
In \(\mathbb{R}, \cap_{n=1}^\infty (1-\frac{1}{n},1+\frac{1}{n})=\{1\}\text{.}\)
Definition 4.39.
A set could be both open and closed (such sets are often called clopen).
Example 4.40.
- In any metric space \(X\text{,}\) \(\emptyset\) and \(X\) are both clopen.
- If \(X=\mathbb{Z}\text{,}\) every subset of \(X\) is clopen, similarly if \(X\) is any set with the discrete metric.
- \(X=[0,1]\cup[2,3]\subseteq \mathbb{R}\) then \([0,1]\) is open and closed in \(X\text{.}\)
Theorem 4.41.
Let M be a metric space and let X\sse M. Then X is open if and only if X^C is closed.
Proof.
(\(\implies\)) Suppose \(U\) is open and \((x_n)_{n=1}^\infty\subseteq X\setminus U\) such that \(\lim_{x\rightarrow\infty}x_n=x\in X\text{.}\) We need to show that \(x\in X\setminus U\text{.}\)
Suppose \(x\in U\text{.}\) Then exists \(\varepsilon>0\) so that \(B_{\varepsilon}(x)\subseteq U\text{.}\) Since \(\lim_{x\rightarrow\infty}x_n=x\text{,}\) there exists \(N\in\mathbb{Z},N\geq 1\) such that for all \(n\geq N, d(x,x_n)<\varepsilon\text{.}\) But then, in particular, \(x_N\in B_\varepsilon(x)\subseteq U\) \(x_n\notin U\text{,}\) a contradiction.
Thus \(x\in X\setminus U\text{.}\)
(\(\impliedby\)) Suppose \(U\) is not open. Then there exists an \(x\in U\) such that for all \(\varepsilon>0,B_\varepsilon(x)\nsubseteq U\text{.}\) For each integer \(n\geq 1\text{,}\) there is an \(x_n\in B_{\frac{1}{n}}(x)\setminus U\text{.}\) Then \((x_n)_{n=1}^\infty\subseteq X\setminus U\) and \(d(x_n,x)\leq \frac{1}{n}\) for all \(n\geq 1\text{.}\) So \(x=\lim_{n\rightarrow\infty}x_n\in U\text{.}\) So \(x\) is a cluster point of \(X\setminus U\text{,}\) but \(x\notin X\setminus U\text{.}\) Hence \(X\setminus U\) is not closed.
Exploration 4.42. Sequential Characterization of Open Sets.
Let \(X\) be a metric space. A set \(U\subseteq X\) is open iff whenever \((x_n)_{n=1}^\infty\subseteq X\) converges and \(\lim_{x\rightarrow\infty}x_n\in U\text{,}\) then there exists \(n\geq 1\) such that for all \(n\geq N,x_n\in U\text{.}\)
