Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.4: Compact Sets

  • Page ID
    22660
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Definition

    Suppose \(T \subset \mathbb{R} .\) If \(A\) is a set, \(U_{\alpha}\) is an open set for every \(\alpha \in A,\) and

    \[T \subset \bigcup_{\alpha \in A} U_{\alpha},\]

    then we call \(\left\{U_{\alpha}: \alpha \in A\right\}\) an open cover of \(T\).

    Example \(\PageIndex{1}\)

    For \(n=3,4,5, \dots,\) let

    \[U_{n}=\left(\frac{1}{n}, \frac{n-1}{n}\right).\]

    Then \(\left\{U_{n}: n=3,4,5, \ldots\right\}\) is an open cover of the open interval \((0,1)\).

    Definition

    Suppose \(\left\{U_{\alpha}: \alpha \in A\right\}\) is an open cover of \(T \subset \mathbb{R} .\) If \(B \subset A\) and

    \[T \subset \bigcup_{\beta \in B} U_{\beta},\]

    then we call \(\left\{U_{\beta}: \beta \in B\right\}\) a subcover of \(\left\{U_{\alpha}: \alpha \in A\right\} .\) If \(B\) is finite, we call \(\left\{U_{\beta}: \beta \in B\right\}\) a finite subcover of \(\left\{U_{\alpha}: \alpha \in A\right\}\).

    Exercise \(\PageIndex{1}\)

    Show that the open cover of \((0,1)\) given in the previous example does not have a finite subcover.

    Definition

    We say a set \(K \subset \mathbb{R}\) is compact if every open cover of \(K\) has a finite sub cover.

    Example \(\PageIndex{2}\)

    As a consequence of the previous exercise, the open interval \((0,1)\) is not compact.

    Exercise \(\PageIndex{2}\)

    Show that every finite subset of \(\mathbb{R}\) is compact.

    Exercise \(\PageIndex{3}\)

    Suppose \(n \in \mathbb{Z}^{+}\) and \(K_{1}, K_{2}, \ldots, K_{n}\) are compact sets. Show that \(\bigcup_{i=1}^{n} K_{i}\) is compact.

    proposition \(\PageIndex{1}\)

    If \(I\) is a closed, bounded interval, then \(I\) is compact.

    Proof

    Let \(a \leq b\) be finite real numbers and \(I=[a, b] .\) Suppose \(\left\{U_{\alpha}: \alpha \in A\right\}\) is an open cover of \(I .\) Let \(\mathcal{O}\) be the set of sets \(\left\{U_{\beta}: \beta \in B\right\}\) with the properties that \(B\) is a finite subset of \(A\) and \(a \in \bigcup_{\beta \in B} U_{\beta} .\) Let

    \[(s-\epsilon, s+\epsilon) \subset U_{\alpha}.\]

    Moreover, there exists a \(\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}\) for which

    \[\left[a, s-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.\]

    But then

    \[\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}\]

    and

    \[\left[a, s+\frac{\epsilon}{2}\right] \subset\left(\bigcup_{\beta \in B} U_{\beta}\right) \cup U_{\alpha},\]

    contradicting the definition of \(s .\) Hence we must have \(s=b .\) Now choose \(U_{\alpha}\) such that \(b \in U_{\alpha} .\) Then, for some \(\epsilon>0\),

    \[(b-\epsilon, b+\epsilon) \subset U_{\alpha}.\]

    Moreover, there exists \(\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}\) such that

    \[\left[a, b-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.\]

    Then

    \[\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}\]

    is a finite subcover of \(I .\) Thus \(I\) is compact. \(\quad\) Q.E.D.

    proposition \(\PageIndex{2}\)

    If \(K\) is a closed, bounded subset of \(\mathbb{R},\) then \(K\) is compact.

    Proof

    Since \(K\) is bounded, there exist finite real numbers \(a\) and \(b\) such that \(K \subset[a, b] .\) Let \(\left\{U_{\alpha}: \alpha \in A\right\}\) be an open cover of \(K .\) Let \(V=\mathbb{R} \backslash K .\) Then

    \[\left\{U_{\alpha}: \alpha \in A\right\} \cup\{V\}\]

    in the latter case, we have

    \[K \subset[a, b] \backslash V \subset \bigcup_{\beta \in B} U_{\beta}.\]

    In either case, we have found a finite subcover of \(\left\{U_{\alpha}: \alpha \in A\right\}\). \(\quad\) Q.E.D.

    Exercise \(\PageIndex{4}\)

    Show that if \(K\) is compact and \(C \subset K\) is closed, then \(C\) is compact.

    proposition \(\PageIndex{3}\)

    If \(K \subset \mathbb{R}\) is compact, then \(K\) is closed.

    Proof

    Suppose \(x\) is a limit point of \(K\) and \(x \notin K .\) For \(n=1,2,3, \ldots,\) let

    \[U_{n}=\left(-\infty, x-\frac{1}{n}\right) \cup\left(x+\frac{1}{n},+\infty\right).\]

    Then

    \[\bigcup_{n=1}^{\infty} U_{n}=(-\infty, x) \cup(x,+\infty) \supset K .\]

    However, for any \(N \in \mathbb{Z}^{+},\) there exists \(a \in K\) with

    \[a \in\left(x-\frac{1}{N}, x+\frac{1}{N}\right),\]

    and hence

    \[a \notin \bigcup_{n=1}^{N} U_{n}=\left(-\infty, x-\frac{1}{N}\right) \cup\left(x+\frac{1}{N},+\infty\right).\]

    Thus the open cover \(\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}\) does not have a finite subcover, contradicting the assumption that \(K\) is compact. \(\quad\) Q.E.D.

    Exercise \(\PageIndex{5}\)

    Suppose that for each \(\alpha\) in some set \(A, K_{\alpha}\) is compact. Show that \(\bigcap_{\alpha \in A} K_{\alpha}\) is compact.

    proposition \(\PageIndex{4}\)

    If \(K \subset \mathbb{R}\) is compact, then \(K\) is bounded.

    Proof

    Suppose \(K\) is not bounded. For \(n=1,2,3, \ldots,\) let \(U_{n}=(-n, n) .\) Then

    \[\bigcup_{n=1}^{\infty} U_{n}=(-\infty, \infty) \supset K .\]

    But, for any integer \(N,\) there exists \(a \in K\) such that \(|a|>N,\) from which it follows that

    \[a \notin \bigcup_{n=1}^{N} U_{n}=(-N, N).\]

    Thus the open cover \(\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}\) does not have a finite subcover, contradicting the assumption that \(K\) is compact. \(\quad\) Q.E.D.

    Taken together, the previous three propositions yield the following fundamental result:

    Theorem \(\PageIndex{5}\)

    A set \(K \subset \mathbb{R}\) is compact if and only if \(K\) is closed and bounded.

    proposition \(\PageIndex{6}\)

    If \(K \subset \mathbb{R}\) is compact and \(\left\{x_{n}\right\}_{n \in I}\) is a sequence with \(x_{n} \in K\) for every \(n \in I,\) then \(\left\{x_{n}\right\}_{n \in I}\) has a convergent subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) with

    \[\lim _{k \rightarrow \infty} x_{n_{k}} \in K .\]

    Proof

    Since \(K\) is bounded, \(\left\{x_{n}\right\}_{n \in I}\) has a convergent subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\). Since \(K\) is closed, we must have \(\lim _{k \rightarrow \infty} x_{n_{k}} \in K\).

    proposition \(\PageIndex{7}\)

    Suppose \(K \subset \mathbb{R}\) is such that whenever \(\left\{x_{n}\right\}_{n \in I}\) is a sequence with \(x_{n} \in K\) for every \(n \in I,\) then \(\left\{x_{n}\right\}_{n \in I}\) has a subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) with \(\lim _{k \rightarrow \infty} x_{n_{k}} \in K .\) Then \(K\) is compact.

    Proof

    Suppose \(K\) is unbounded. Then we may construct a sequence \(\left\{x_{n}\right\}_{n=1}^{\infty}\) such that \(x_{n} \in K\) and \(\left|x_{n}\right|>n\) for \(n=1,2,3, \ldots\) Hence the only possible subsequential limits of \(\left\{x_{n}\right\}_{n=1}^{\infty}\) would be \(-\infty\) and \(+\infty,\) contradicting our assumptions. Thus \(K\) must be bounded.

    Now suppose \(\left\{x_{n}\right\}_{n \in I}\) is a convergent sequence with \(x_{n} \in K\) for all \(n \in I .\) If \(L=\lim _{n \rightarrow \infty} x_{n},\) then \(L\) is the only subsequential limit of \(\left\{x_{n}\right\}_{n \in I} .\) Hence, by the assumptions of the proposition, \(L \in K .\) Hence \(K\) is closed.

    Since \(K\) is both closed and bounded, it is compact. \(\quad\) Q.E.D.

    Exercise \(\PageIndex{6}\)

    Show that a set \(K \subset \mathbb{R}\) is compact if and only if every infinite subset of \(K\) has a limit point in \(K\).

    Exercise \(\PageIndex{7}\)

    Show that if \(K\) is compact, then \(\sup K \in K\) and \(\inf K \in K\).

    Theorem \(\PageIndex{8}\)

    Given a set \(K \subset \mathbb{R},\) the following are equivalent:

    1. Every open cover of \(K\) has a finite subcover.

    2. Every sequence in \(K\) has a subsequential limit in \(K\).

    3. Every infinite subset of \(K\) has a limit point in \(K\).

    Exercise \(\PageIndex{8}\)

    Suppose \(K_{1}, K_{2}, K_{3}, \ldots\) are nonempty compact sets with

    \[K_{n+1} \subset K_{n}\]

    for \(n=1,2,3, \ldots\) Show that

    \[\bigcap_{n=1}^{\infty} K_{n}\]

    is nonempty.

    Exercise \(\PageIndex{9}\)

    We say a collection of sets \(\left\{D_{\alpha}: \alpha \in A\right\}\) has the finite intersection property if for every finite set \(B \subset A\),

    \[\bigcap_{\alpha \in B} D_{\alpha} \neq \emptyset .\]

    Show that a set \(K \subset \mathbb{R}\) is compact if and only for any collection

    \[\left\{E_{\alpha}: \alpha \in A, E_{\alpha}=C_{\alpha} \cap K \text { where } C_{\alpha} \subset \mathbb{R} \text { is closed }\right\}\]

    which has the finite intersection property we have

    \[\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset .\]

    • Was this article helpful?