4.3: Closed Sets
- Page ID
- 22659
We call a point \(x \in \mathbb{R}\) a limit point of a set \(A \subset \mathbb{R}\) if for every \(\epsilon>0\) there exists \(a \in A, a \neq x,\) such that \(a \in(x-\epsilon, x+\epsilon)\).
Suppose \(A \subset \mathbb{R} .\) We call a point \(a \in A\) an isolated point of \(A\) if there exists an \(\epsilon>0\) such that \[A \cap(a-\epsilon, a+\epsilon)=\{a\}.\]
Identify the limit points and isolated points of the following sets:
a. \([-1,1]\),
b. \((-1,1)\),
c. \(\left\{\frac{1}{n}: n \in \mathbb{Z}^{+}\right\}\),
d. \(\mathbb{Z}\),
e. \(\mathbb{Q}\).
Suppose \(x\) is a limit point of the set \(A .\) Show that for every \(\epsilon>0,\) the set \((x-\epsilon, x+\epsilon) \cap A\) is infinite.
We let \(A^{\prime}\) denote the set of limit points of a set \(A .\)
Given a set \(A \subset \mathbb{R},\) we call the set \(\bar{A}=A \cup A^{\prime}\) the closure of \(A\).
We call a set \(C \subset \mathbb{R}\) closed if \(C=\bar{C}\).
If \(A \subset \mathbb{R},\) then \(\bar{A}\) is closed.
- Proof
-
Suppose \(x\) is a limit point of \(\bar{A} .\) We we will show that \(x\) is a limit point of \(A,\) and hence \(x \in \bar{A} .\) Now for any \(\epsilon>0,\) there exists \(a \in \bar{A}, a \neq x,\) such that
\[a \in\left(x-\frac{\epsilon}{2}, x+\frac{\epsilon}{2}\right).\]
If \(a \in A,\) let \(b=a .\) If \(a \notin A,\) then \(a\) is a limit point of \(A,\) so there exists \(b \in A,\) \(b \neq a\) and \(b \neq x,\) such that
\[b \in\left(a-\frac{\epsilon}{2}, a+\frac{\epsilon}{2}\right).\]
In either case
\[|x-b| \leq|x-a|+|a-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .\]
Hence \(x \in A^{\prime},\) and so \(\bar{A}\) is closed. \(\quad\) Q.E.D.
A set \(C \subset \mathbb{R}\) is closed if and only if for every convergent sequence \(\left\{a_{k}\right\}_{k \in K}\) with \(a_{k} \in C\) for all \(k \in K\),
\[\lim _{k \rightarrow \infty} a_{k} \in C.\]
- Proof
-
Suppose \(C\) is closed and \(\left\{a_{k}\right\}_{k \in K}\) is a convergent sequence with \(a_{k} \in C\) for all \(k \in K .\) Let \(x=\lim _{k \rightarrow \infty} a_{k} .\) If \(x=a_{k}\) for some integer \(k,\) then \(x \in C .\) Otherwise, for every \(\epsilon>0,\) there exists an integer \(N\) such that \(\left|a_{N}-x\right|<\epsilon\). Hence \(a_{N} \neq x\) and
\[a_{N} \in(x-\epsilon, x+\epsilon).\]
Thus \(x\) is a limit point of \(C,\) and so \(x \in C\) since \(C\) is closed.
Now suppose that for every convergent sequence \(\left\{a_{k}\right\}_{k \in K}\) with \(a_{k} \in C\) for all \(k \in K, \lim _{k \rightarrow \infty} a_{k} \in C .\) Let \(x\) be a limit point of \(C .\) For \(k=1,2,3, \ldots,\) choose \(a_{k} \in C\) such that \(a_{k} \in\left(x-\frac{1}{k}, x+\frac{1}{k}\right) .\) Then clearly
\[\boldsymbol{x}=\lim _{k \rightarrow \infty} a_{k},\]
so \(x \in C .\) Thus \(C\) is closed. \(\quad\) Q.E.D.
Show that every closed interval \(I\) is a closed set.
Suppose \(A\) is a set and, for each \(\alpha \in A, C_{\alpha}\) is a closed set. Then
\[\bigcap_{\alpha \in A} C_{\alpha}\]
is a closed set.
- Proof
-
Suppose \(x\) is a limit point of \(\bigcap_{\alpha \in A} C_{\alpha} .\) Then for any \(\epsilon>0,\) there exists \(y \in \bigcap_{\alpha \in A} C_{\alpha}\) such that \(y \neq x\) and \(y \in(x-\epsilon, x+\epsilon) .\) But then for any \(\alpha \in A,\) \(y \in C_{\alpha},\) so \(x\) is a limit point of \(C_{\alpha}\). Since \(C_{\alpha}\) is closed, it follows that \(x \in C_{\alpha}\) for every \(\alpha \in A .\) Thus \(x \in \bigcap_{\alpha \in A} C_{\alpha}\) and \(\bigcap_{\alpha \in A} C_{\alpha}\) is closed. \(\quad\) Q.E.D.
Suppose \(C_{1}, C_{2}, \ldots, C_{n}\) is a finite collection of closed sets. Then
\[\bigcup_{i=1}^{n} C_{i}\]
is closed.
- Proof
-
Suppose \(\left\{a_{k}\right\}_{k \in K}\) is a convergent sequence with \(a_{k} \in \bigcup_{i=1}^{n} C_{i}\) for every \(k \in K .\) Let \(L=\lim _{k \rightarrow \infty} a_{k} .\) Since \(K\) is an infinite set, there must an integer \(m\) and a subsequence \(\left\{a_{n_{j}}\right\}_{j=1}^{\infty}\) such that \(a_{n_{j}} \in C_{m}\) for \(j=1,2, \ldots\). Since every subsequence of \(\left\{a_{k}\right\}_{k \in K}\) converges to \(L,\left\{a_{n_{j}}\right\}_{j=1}^{\infty}\) must converge to \(L .\) Since \(C_{m}\) is closed,
\[L=\lim _{j \rightarrow \infty} a_{n_{j}} \in C_{m} \subset \bigcup_{i=1}^{n} C_{i}.\]
Thus \(\bigcup_{i=1}^{n} C_{i}\) is closed. \(\quad\) Q.E.D.
Note that both \(\mathbb{R}\) and \(\emptyset\) satisfy the definition of a closed set.
A set \(C \subset \mathbb{R}\) is closed if and only if \(\mathbb{R} \backslash C\) is open.
- Proof
-
Assume \(C\) is closed and let \(U=\mathbb{R} \backslash C .\) If \(C=\mathbb{R},\) then \(U=\emptyset,\) which is open; if \(C=\emptyset,\) then \(U=\mathbb{R},\) which is open. So we may assume both \(C\) and \(U\) are nonempty. Let \(x \in U .\) Then \(x\) is not a limit point of \(C,\) so there exists an \(\epsilon>0\) such that
\[(x-\epsilon, x+\epsilon) \cap C=\emptyset.\]
Thus
\[(x-\epsilon, x+\epsilon) \subset U,\]
so \(U\) is open.
Now suppose \(U=\mathbb{R} \backslash C\) is open. If \(U=\mathbb{R},\) then \(C=\emptyset,\) which is closed; if \(U=\emptyset,\) then \(C=\mathbb{R},\) which is closed. So we may assume both \(U\) and \(C\) are nonempty. Let \(x\) be a limit point of \(C .\) Then, for every \(\epsilon>0\),
\[(x-\epsilon, x+\epsilon) \cap C \neq \emptyset .\]
Hence there does not exist \(\epsilon>0\) such that
\[(x-\epsilon, x+\epsilon) \subset U.\]
Thus \(x \notin U,\) so \(x \in C\) and \(C\) is closed. \(\quad\) Q.E.D.
For \(n=1,2,3, \ldots,\) let \(I_{n}=\left(-\frac{1}{n}, \frac{n+1}{n}\right) .\) Is
\[\bigcap_{n=1}^{\infty} I_{n}\]
open or closed?
For \(n=3,4,5, \ldots,\) let \(I_{n}=\left[\frac{1}{n}, \frac{n-1}{n}\right] .\) Is
\[\bigcup_{n=3}^{\infty} I_{n}\]
open or closed?
Suppose, for \(n=1,2,3, \ldots,\) the intervals \(I_{n}=\left[a_{n}, b_{n}\right]\) are such that \(I_{n+1} \subset I_{n} .\) If \(a=\sup \left\{a_{n}: n \in \mathbb{Z}^{+}\right\}\) and \(b=\inf \left\{b_{n}: n \in \mathbb{Z}^{+}\right\},\) show that
\[\bigcap_{n=1}^{\infty} I_{n}=[a, b].\]
Find a sequence \(I_{n}, n=1,2,3, \ldots,\) of closed intervals such that \(I_{n+1} \subset I_{n}\) for \(n=1,2,3, \ldots\) and
\[\bigcap_{n=1}^{\infty} I_{n}=\emptyset.\]
Find a sequence \(I_{n}, n=1,2,3, \ldots,\) of bounded, open intervals such that \(I_{n+1} \subset I_{n}\) for \(n=1,2,3, \ldots\) and
\[\bigcap_{n=1}^{\infty} I_{n}=\emptyset .\]
Suppose \(A_{i} \subset \mathbb{R}, i=1,2, \ldots, n,\) and let \(B=\bigcup_{i=1}^{n} A_{i} .\) Show that
\[\overline{B}=\bigcup_{i=1}^{n} \overline{A_{i}}.\]
Suppose \(A_{i} \subset \mathbb{R}, i \in \mathbb{Z}^{+},\) and let
\[B=\bigcup_{i=1}^{\infty} A_{i}.\]
Show that
\[\bigcup_{i=1}^{\infty} \overline{A_{i}} \subset \overline{B} .\]
Find an example for which
\[\overline{B} \neq \bigcup_{i=1}^{\infty} \overline{A_{i}}.\]
Suppose \(U \subset \mathbb{R}\) is a nonempty open set. For each \(x \in U,\) let
\[J_{x}=\bigcup(x-\epsilon, x+\delta),\]
where the union is taken over all \(\epsilon>0\) and \(\delta>0\) such that \((x-\epsilon, x+\delta) \subset U\).
a. Show that for every \(x, y \in U,\) either \(J_{x} \cap J_{y}=\emptyset\) or \(J_{x}=J_{y}\).
b. Show that
\[U=\bigcup_{x \in B} J_{x},\]
where \(B \subset U\) is either finite or countable.