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Mathematics LibreTexts

4.3: Closed Sets

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    22659
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    Definition: limit points

    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 \(a \in(x-\epsilon, x+\epsilon)\).

    Definition: isolated point

    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 \[A \cap(a-\epsilon, a+\epsilon)=\{a\}.\]

    Exercise \(\PageIndex{1}\)

    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. \(\mathbb{Q}\).

    Exercise \(\PageIndex{2}\)

    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 \(A .\)

    Definition: closures

    Given a set \(A \subset \mathbb{R},\) we call the set \(\bar{A}=A \cup A^{\prime}\) the closure of \(A\).

    Definition: Closed sets

    We call a set \(C \subset \mathbb{R}\) closed if \(C=\bar{C}\).

    Proposition \(\PageIndex{1}\)

    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.

    Proposition \(\PageIndex{2}\)

    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.

    Exercise \(\PageIndex{3}\)

    Show that every closed interval \(I\) is a closed set.

    Proposition \(\PageIndex{3}\)

    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.

    Proposition \(\PageIndex{4}\)

    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.

    Proposition \(\PageIndex{5}\)

    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.

    Exercise \(\PageIndex{4}\)

    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?

    Exercise \(\PageIndex{5}\)

    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?

    Exercise \(\PageIndex{6}\)

    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

    \[\bigcap_{n=1}^{\infty} I_{n}=[a, b].\]

    Exercise \(\PageIndex{7}\)

    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

    \[\bigcap_{n=1}^{\infty} I_{n}=\emptyset.\]

    Exercise \(\PageIndex{8}\)

    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

    \[\bigcap_{n=1}^{\infty} I_{n}=\emptyset .\]

    Exercise \(\PageIndex{9}\)

    Suppose \(A_{i} \subset \mathbb{R}, i=1,2, \ldots, n,\) and let \(B=\bigcup_{i=1}^{n} A_{i} .\) Show that

    \[\overline{B}=\bigcup_{i=1}^{n} \overline{A_{i}}.\]

    Exercise \(\PageIndex{10}\)

    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

    \[\overline{B} \neq \bigcup_{i=1}^{\infty} \overline{A_{i}}.\]

    Exercise \(\PageIndex{11}\)

    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.

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