Compactness

Sequential compactness

Definition (Sequentially compact set)

\(C \subseteq X\) is sequentially compact if every sequence \(\{ x_{n}\}_{n=1}^{\infty} \subseteq C\) has a subsequence converging to a point of \(C\).

Lemma (Characterisation by intersection of the closures of the tails)

Let \((X, d)\) be a metric space and \(C \subseteq X\). The following are equivalent:

1. \(C\) is sequentially compact;

2. for every \(\{ x_{n}\}_{n=1}^{\infty} \subseteq C\),

   \[C \cap \left( \bigcap_{k=1}^{\infty} \overline{ \{ x_{m} : m \geq k\}} \right) \neq \emptyset.\]

Proof: \((1) \implies (2)\): If \(\{ x_{n_{k}}\}_{k=1}^{\infty}\) is a subsequence converging to \(x_{0} \in C\), then given \(\varepsilon > 0\) there exists \(k_{0}\) such that \(k \geq k_{0}\) implies \(d(x_{n_{k}}, x_{0}) < \varepsilon\). Since \(n_{k} \geq k\), it follows that \(B(x_{0}, \varepsilon) \cap \{ x_{m} : m \geq k\} \neq \emptyset\) for every \(k \geq 1\), so \(x_{0} \in \overline{ \{ x_{m} : m \geq k\}}\) for every \(k\).

\((2) \implies (1)\): Let \(x_{0} \in C \cap \bigcap_{k=1}^{\infty} \overline{ \{ x_{m} : m \geq k\}}\). We iteratively construct a subsequence converging to \(x_{0}\): we take \(x_{n_{1}} \in B(x_{0}, 1) \cap \{ x_{m} : m \geq 1\}\), then \(x_{n_{2}} \in B(x_{0}, \tfrac{1}{2}) \cap \{ x_{m} : m \geq n_{1} + 1\}\), and in general \(x_{n_{k+1}} \in B(x_{0}, \tfrac{1}{k}) \cap \{ x_{m} : m \geq n_{k} + 1\}\). Then \(x_{n_{k}} \to x_{0}\).

Lemma (Sequentially compact implies closed and bounded)

Let \(C \subseteq X\) be sequentially compact. Then \(C\) is closed and bounded.

Proof: Closed: Let \(x \in \overline{C}\). There exists \(\{x_{n}\}_{n=1}^{\infty} \subseteq C\) with \(x_{n} \to x\). By sequential compactness there exists a subsequence \(\{x_{n_{k}}\}_{k=1}^{\infty}\) converging to a point of \(C\). By uniqueness of the limit, \(x \in C\), so \(\overline{C} \subseteq C\).

Bounded: If \(C\) were not bounded, then, fixing \(x_{0} \in X\), we could choose \(x_{n} \in C\) with \(d(x_{0}, x_{n}) \geq n\). Every subsequence \(\{x_{n_{k}}\}\) would satisfy \(d(x_{0}, x_{n_{k}}) \to \infty\) and hence could not be convergent, contradicting sequential compactness.

Open covers and compactness

Definition (Open cover)

A collection \(\mathcal{U} = \{ U_{\alpha} : \alpha \in \Lambda\}\) of open sets is a cover (or covering) of \(A \subseteq X\) if

\[A \subseteq \bigcup_{\alpha \in \Lambda} U_{\alpha}.\]

Lemma (Technical covering lemma for sequentially compact sets (Lebesgue))

Let \(C \subseteq X\) be sequentially compact and \(\mathcal{U}\) an open cover of \(C\). Then there exists \(\varepsilon > 0\) such that for every \(x \in C\) there exists \(U \in \mathcal{U}\) with \(B(x, \varepsilon) \subseteq U\).

Proof: By contradiction: suppose that for every \(\varepsilon > 0\) there exists \(x_{\varepsilon} \in C\) such that \(B(x_{\varepsilon}, \varepsilon) \not\subseteq U\) for any \(U \in \mathcal{U}\). In particular, for \(\varepsilon = 1/n\) there exists \(x_{n} \in C\) with \(B(x_{n}, 1/n) \not\subseteq U_{\alpha}\) for every \(\alpha \in \Lambda\).

By sequential compactness there exist \(\{x_{n_{k}}\}_{k=1}^{\infty}\) and \(x_{0} \in C\) such that \(x_{n_{k}} \to x_{0}\). Since \(\mathcal{U}\) covers \(C\), there exists \(U_{\alpha_{0}} \in \mathcal{U}\) with \(x_{0} \in U_{\alpha_{0}}\) and \(\varepsilon > 0\) such that \(B(x_{0}, \varepsilon) \subseteq U_{\alpha_{0}}\). Take \(k_{0}\) such that \(k \geq k_{0}\) implies \(d(x_{n_{k}}, x_{0}) < \varepsilon/2\) and \(1/n_{k} < \varepsilon/2\). Then, for \(y \in B(x_{n_{k}}, 1/n_{k})\),

\[d(y, x_{0}) \leq d(y, x_{n_{k}}) + d(x_{n_{k}}, x_{0}) < \tfrac{1}{n_{k}} + \tfrac{\varepsilon}{2} < \varepsilon,\]

so \(B(x_{n_{k}}, 1/n_{k}) \subseteq B(x_{0}, \varepsilon) \subseteq U_{\alpha_{0}}\), which contradicts the construction.

Definition (Compact set)

\(C \subseteq X\) is compact if, given an open cover \(\mathcal{U} = \{U_{\alpha} : \alpha \in \Lambda\}\) of \(C\), there exist \(U_{\alpha_{1}}, \dots, U_{\alpha_{m}} \in \mathcal{U}\) such that

\[C \subseteq \bigcup_{k=1}^{m} U_{\alpha_{k}}.\]

That is, every open cover admits a finite subcover.

Equivalence of the two notions of compactness

Lemma (Sequential compactness is equivalent to compactness)

For \(C \subseteq X\) the following are equivalent:

1. \(C\) is sequentially compact;
2. \(C\) is compact.

Proof: \((1) \implies (2)\): Let \(\mathcal{U}\) be an open cover of \(C\). By the technical lemma above there exists \(\varepsilon > 0\) such that for each \(x \in C\) some ball \(B(x, \varepsilon)\) is contained in some \(U_{\alpha}\). Take \(x_{1} \in C\) and \(\alpha_{1}\) with \(B(x_{1}, \varepsilon) \subseteq U_{\alpha_{1}}\). If \(C \subseteq B(x_{1}, \varepsilon)\) we already have a finite subcover; if not, we choose \(x_{2} \in C \setminus B(x_{1}, \varepsilon)\) and \(U_{\alpha_{2}}\) with \(B(x_{2}, \varepsilon) \subseteq U_{\alpha_{2}}\). Iterating, if at some step \(C \subseteq \bigcup_{i=1}^{m} B(x_{i}, \varepsilon) \subseteq \bigcup_{i=1}^{m} U_{\alpha_{i}}\), we are done. Otherwise, we obtain \(\{x_{k}\}_{k=1}^{\infty} \subseteq C\) with \(d(x_{i}, x_{j}) \geq \varepsilon\) for \(i \neq j\). This sequence admits no convergent subsequence: every convergent subsequence would be Cauchy, so its terms would eventually be at distance \(< \varepsilon\), contradicting \(d(x_{i}, x_{j}) \geq \varepsilon\). This contradicts sequential compactness.

\((2) \implies (1)\): Let \(\{x_{m}\}_{m=1}^{\infty} \subseteq C\) and suppose, by contradiction, that it admits no subsequence converging in \(C\). Define \(U_{n} = X \setminus \overline{\{x_{m} : m \geq n\}}\). Then \(U_{n} \subseteq U_{n+1}\) and \(X \setminus \bigcup_{n=1}^{\infty} U_{n} = \bigcap_{n=1}^{\infty} \overline{\{x_{m} : m \geq n\}}\). Since \(\{x_{m}\}\) has no subsequence converging in \(C\), \(C \cap \bigcap_{k=1}^{\infty} \overline{\{x_{m} : m \geq k\}} = \emptyset\), so \(C \subseteq \bigcup_{n=1}^{\infty} U_{n}\). By compactness there exists a finite subcover; by monotonicity of \(\{U_{n}\}\), there exists \(m_{0}\) with \(C \subseteq U_{m_{0}}\). But then \(\{x_{k} : k \geq m_{0}\} \subseteq C \subseteq X \setminus \overline{\{x_{k} : k \geq m_{0}\}}\), a contradiction.

Compactness and continuous functions

Lemma (The continuous image of a compact set is compact)

Let \(f : X \to Y\) be continuous and \(K \subseteq X\) compact. Then \(f(K)\) is compact in \(Y\).

Proof: Let \(\mathcal{U} = \{ U_{\alpha} : \alpha \in \Lambda_{1}\}\) be an open cover of \(f(K)\). Then \(K \subseteq f^{-1}\!\left( \bigcup_{\alpha \in \Lambda_{1}} U_{\alpha}\right) = \bigcup_{\alpha \in \Lambda_{1}} f^{-1}(U_{\alpha})\), and each \(f^{-1}(U_{\alpha})\) is open in \(X\). By compactness of \(K\), there exist \(\alpha_{1}, \dots, \alpha_{m}\) such that

\[K \subseteq \bigcup_{i=1}^{m} f^{-1}(U_{\alpha_{i}}),\]

from which \(f(K) \subseteq \bigcup_{i=1}^{m} U_{\alpha_{i}}\).

Theorem (Characterisation by the finite intersection property)

Let \((X, d)\) be a metric space. The following are equivalent:

1. \(X\) is compact;
2. if \(\mathcal{F} = \{ F_{\alpha} : \alpha \in \Lambda\}\) is a family of closed sets such that \(\bigcap_{\alpha \in A} F_{\alpha} \neq \emptyset\) for every finite \(A \subseteq \Lambda\), then \(\bigcap_{\alpha \in \Lambda} F_{\alpha} \neq \emptyset\).

Proof: For both directions we use that \(\bigcap_{\alpha \in \Lambda} F_{\alpha} = \emptyset\) if and only if \(X = \bigcup_{\alpha \in \Lambda} (X \setminus F_{\alpha})\), that is, \(\{X \setminus F_{\alpha}\}_{\alpha}\) is an open cover of \(X\).

If \(X\) is compact and the finite intersection property holds but \(\bigcap_{\alpha} F_{\alpha} = \emptyset\), then there would exist \(\alpha_{1}, \dots, \alpha_{m}\) with \(X = \bigcup_{i=1}^{m} (X \setminus F_{\alpha_{i}})\), i.e. \(\bigcap_{i=1}^{m} F_{\alpha_{i}} = \emptyset\), a contradiction.

Conversely, given an open cover \(\{ U_{\alpha}\}\) with no finite subcover, \(\{X \setminus U_{\alpha}\}\) is a family of closed sets with the finite intersection property but with empty total intersection.

Compactness in Rd

Note (Boxes in \(\mathbb{R}^{d}\) are sequentially compact)

By the Bolzano-Weierstrass theorem, \(C = [a_{1}, b_{1}] \times \dots \times [a_{d}, b_{d}]\) with \(a_{i} \leq b_{i}\) is sequentially compact. Moreover every compact set is closed and bounded.

Theorem (Heine-Borel in \(\mathbb{R}^{d}\))

Let \(C \subseteq \mathbb{R}^{d}\). Then \(C\) is compact with respect to the Euclidean norm if and only if \(C\) is closed and bounded.

Proof: \((\implies)\): Every compact set is closed and bounded by the general lemma.

\((\impliedby)\): Let \(F \subseteq \mathbb{R}^{d}\) be closed and bounded. There exists \(n\) such that \(F \subseteq C_{n} = [-n, n]^{d}\). Let \(\{x_{m}\}_{m=1}^{\infty} \subseteq F\). Since \(C_{n}\) is sequentially compact, there exists a subsequence \(\{x_{m_{k}}\}_{k=1}^{\infty}\) converging to \(x_{0} \in C_{n}\). Since \(\{x_{m_{k}}\} \subseteq F\) and \(F\) is closed, \(x_{0} \in F\). Thus \(F\) is sequentially compact and, by the equivalence, compact.

Theorem (Extreme values)

Let \(K \subseteq X\) be compact and \(f : X \to \mathbb{R}\) continuous. Then \(f(K)\) is compact and, in particular, \(f\) attains its infimum and its supremum on \(K\).

Proof: \(f(K)\) is compact, hence closed and bounded in \(\mathbb{R}\). Let \(a = \inf_{x \in K} f(x)\) and \(b = \sup_{x \in K} f(x)\). For each \(n \in \mathbb{N}^{\ast}\) there exists \(x_{n} \in K\) with \(a \leq f(x_{n}) \leq a + 1/n\). By compactness of \(K\), there exists a subsequence \(\{x_{n_{k}}\}\) converging to \(y_{0} \in K\), and by continuity \(f(x_{n_{k}}) \to f(y_{0})\); but \(f(x_{n_{k}}) \to a\), hence \(f(y_{0}) = a\). Analogously, \(b\) is attained.