Real Analysis

Compact Sets

Set Covers

An open cover, or cover, of a set \(E\) in metric space \(X\) is a collection of open sets, denoted \(\{G_\alpha\}\), whose union covers, or contains, \(E\).

A subcover of \(\{G_\alpha\}\) is a subcollection \(\{G_{\alpha_\gamma}\}\) that still covers \(E\).

For example,

  1. In \(\mathbb{R}\), we have \(\left[\frac{1}{2}, 1\right)\) has cover \(\{V_n\}_{n = 3}^{\infty}\) where \(V_n = \left(\frac{1}{n}, 1 - \frac{1}{n}\right)\). It also has cover \(\{(0, 2)\}\). It also has cover \(\{W_x\}_{x \in \left(\frac{1}{2}, 1\right)}\), where \(W_x = N_\frac{1}{10}(x)\). \(\{V_n\}_{n = 3}^{\infty}\) has subcovers \(\{V_n\}_{n = 22}^{\infty}\) and \(\{V_n\}_{n = 1000000}^{\infty}\). \(\{(0, 2)\}\) has only one subcover \(\{(0, 2)\}\). \(\{W_x\}_{x \in \left(\frac{1}{2}, 1\right)}\) has a subcover \(\left\{W_\frac{5}{10}, W_\frac{6}{10}, W_\frac{7}{10}, W_\frac{8}{10}, W_\frac{8}{10}, W_\frac{9}{10}\right\}\).
  2. \([0, 1] \in \mathbb{R}\) has a cover by \(\{V_n\} \cup \{W_0, W_1\}\). A finite subcover is \(\{W_0, W_1, V_{11}\}\).

Compact Sets

A set \(K\) is compact (in \(X\)) if every open cover of \(K\) contains a finite subcover.

So, \(K\) is not compact means there exists an open cover of \(K\) with no finite subcover.

For example,

  1. \(\left[\frac{1}{2}, 1\right)\) is not compact (see \(\{V_n\}\)).
  2. \(\mathbb{Z}\) in \(\mathbb{R}\) is not compact. One witness cover is \(\{U_i\}_{i \in \mathbb{Z}}\), where \(U_i = \left(i - \frac{1}{10}, i + \frac{1}{10}\right)\).

Theorem. Finite sets are compact.

Proof. Consider some open cover \(\{G_\alpha\}\) covering \(x_1, x_2, \dots, x_n\). For all \(x_i\), choose \(G_{\alpha_i}\). Then \(\{G_{\alpha_i}\}_{i = 1}^n\) covers the set. QED.

A set \(K\) is bounded if there is some ball \(N_r(x)\) for some \(x \in X\) such that \(K \subset N_r(x)\).

Theorem. Compact sets are bounded.

Proof. Consider a compact set \(K\). Let \(B(x) = N_1(x)\), i.e. \(B(x)\) are balls of radius \(1\). Consider a collection \(\{B(x)\}_{x \in K}\), which is an open cover of \(K\). By compactness of \(K\), there exists a finite subcover \(\{B(x_i)\}_{i = 1}^n\). Let \(R = max_{1 \leq i \leq n}\{d(x_1, x_i)\}\); this maximum exists because the set \(\{x_1, \dots, x_n\}\) is finite. Then \(N_{R + 2}(x_1)\) contains all of \(K\). QED.

Relative Open Sets

If \(Y \subset X\), where \(X\) is a metric space, \(Y\) is said to inherit a metric from \(X\).

A set \(U\) is open in \(Y\) (or open relative to \(Y\)) if every point of \(U\) is an interior point of \(U\). Here, the notion of interior is using the neighborhoods in \(Y\).

Theorem. Suppose \(E \subset Y \subset X\). Then \(E\) is open in \(Y\) if and only if \(E = Y \cap G\) for some \(G\) open in \(X\).

Compactness is Intrinsic

Theorem. Suppose \(K \subset Y \subset X\), where \(X\) is a metric space. Then \(K\) is compact in \(Y\) if and only if \(K\) is compact in \(X\).

Proof.

(\(\Rightarrow\)): Assume \(K\) is compact in \(Y\). Consider any open cover \(\{U_\alpha\}\) of \(K\) in \(X\). Let \(V_\alpha = U_\alpha \cap Y\). Then \(\{V_\alpha\}\) covers \(K\) in \(Y\). Since \(K\) is compact in \(Y\), there exists a finite subcover \(\{V_{\alpha_1}, \dots, V_{\alpha_n}\}\) in \(Y\). Then \(\{U_{\alpha_1}, \dots, U_{\alpha_n}\}\) is a finite subcover of \(K\) in \(X\), as desired.

(\(\Leftarrow\)): Assume \(K\) is compact in \(X\). Consider any open cover \(\{V_\alpha\}\) of \(K\) in \(Y\). By above theorem, there exists open \(U_\alpha\) such that \(V_\alpha = U_\alpha \cap Y\). Then \(\{U_\alpha\}\) covers \(K\) in \(X\). Since \(K\) is compact in \(X\), there exists a finite subcover \(\{U_{\alpha_1}, \dots, U_{\alpha_n}\}\). Then \(\{V_{\alpha_1}, \dots, V_{\alpha_n}\}\) is a finite subcover of \(K\) in \(Y\), as desired. QED.

Relationship between Compact Sets and Closed Sets

Theorem. Compact sets are closed.

Proof. Let \(K\) be compact. Consider \(p \notin K\). We’ll show \(p\) has a neighborhood that does not intersect \(K\) (or \(p\) is not a limit point of \(K\)). For any \(q \in K\), let \(V_q = N_\frac{r}{2}(q)\) and \(U_q = N_\frac{r}{2}(p)\), where \(r = d(p, q)\). Notice \(\{V_q\}\) is an open cover of \(K\). By compactness of \(K\), there exists a finite subcover \(\{V_{q_1}, \dots, V_{q_n}\}\). Let \(W = \bigcap_{i = 1}^n U_{q_i}\). So \(W\) is an intersection of finitely many open sets. Hence, \(W\) is open (it is a ball of radius \(\min \left\{\frac{d(p, q_i)}{2}\right\}\)). We have \(W \cap V_{q_i} = \emptyset\) for all \(q_i\) because \(W \subset V_{q_i}\) and \(V_{q_i} \cap U_{q_i} = \emptyset\). So \(W\) is the desired neighborhood. QED.

For example,

  1. \((0, 1)\) in \(\mathbb{R}\) is not compact because it’s not closed.
  2. \(\mathbb{R}\) in \(\mathbb{R}\) is not compact although it’s closed because it’s not bounded.

Theorem. A closed subset \(B\) of a compact set \(K\) is compact.

Proof. Consider any open cover \(\{U_\alpha\}\) of \(B\). Since \(B\) is closed, \(B^c\) is open. So \(B^c \cup \{U_\alpha\}\) is an open cover of \(K\). Since \(K\) is compact, there exists a finite subcover \(\{U_{\alpha_1}, \dots, U_{\alpha_n}, B^c\}\) of \(K\). Since \(B^c \cap B = \emptyset\), \(\{U_{\alpha_1}, \dots, U_{\alpha_n}\}\) covers \(B\) and it is a finite subcover of the original cover. QED.

Corollary. Suppose set \(B\) is closed and set \(K\) is compact. Then \(B \cap K\) is compact.

Consider intervals \(I_n = [a_n, b_n]\). They are nested means if \(m > n\) then \(a_n \leq a_m \leq b_m \leq b_n\).

Theorem. Nested closed intervals in \(\mathbb{R}\) are not empty (in \(\mathbb{R}^k\), nested closed \(k\)-cells are not empty).

Proof. Let \(x = \sup\{a_i\}\). It exists because \(a_i\)’s are bounded by \(b_1\). Since \(x\) is a supremum, \(x \geq a_i\) for all \(i\). We have \(x \leq b_i\) for all \(i\) since \(b_i\) is an upper bound of all \(a_i\)’s (due to nestedness). QED.

Aside: Proof that \(\mathbb{R}\) is uncountable. Suppose \(\mathbb{R}\) is countable, i.e. \(\mathbb{R} = \{x_1, x_2, x_3, \dots\}\). Choose closed intervals \(I_1\) that misses \(x_1\), \(I_2 \subset I_1\) that misses \(x_1\) and \(x_2\), \(I_3 \subset I_2\) that misses \(x_1\), \(x_2\), and \(x_3\), and so on. So, \(I_n\) is a nested closed interval. By previous theorem, there exists a point in \(\bigcap I_n\) that is not one of the \(x_i\)’s and is thus not on the original listing. QED.

Theorem. Any closed interval \([a, b]\) is compact (in \(\mathbb{R}\)). This is also true for \(k\)-cells in \(\mathbb{R}^k\).

Proof (by contradiction). Suppose there is a closed interval \([a, b]\) that is not compact. Then there exists an open cover \(\{U_\alpha\}\) for \([a, b]\) which has no finite subcover. Then \(\{U_\alpha\}\) covers both \([a, c_1]\) and \([c_1, b]\), at least one of which has no finite subcover of \(\{U_\alpha\}\). Without loss of generality, assume \(I_1 = [a, c_1]\) has no finite subcover. Subdivide \(I_1\) again using \(c_2\) which is the half way point of \(a\) and \(c_1\). Note that at least one of \([a, c_2]\) and \([c_2, c_1]\) has no finite subcover. Continue and we’ll obtain \(I_1 \supset I_2 \supset I_3 \supset \dots\) of nested closed intervals, each halves at each step and has no finite subcover of \(\{U_\alpha\}\). By the nested interval theorem, there exists \(x \in I_n\) for all \(n\). Then \(x\) is in some \(U_\widehat{\alpha}\) of the cover \(\{U_\alpha\}\). Since \(U_\widehat{\alpha}\) is open, there exists an \(r > 0\) such that \(N_r(x) \subset U_\widehat{\alpha}\). Since the intervals halve at each step, some \(I_n\) is contained in \(N_r(x)\). This means that \(G_\widehat{\alpha}\) covers \(I_n\), which contradicts the fact that it has no finite subcover. QED.

The Heine-Borel Theorem

Theorem. In \(\mathbb{R}\) (or in \(\mathbb{R}^n\)), \(K\) is compact if and only if \(K\) is both closed and bounded.

Proof.

(\(\Rightarrow\)): already done.

(\(\Leftarrow\)): (it is not true in arbitrary metric spaces) Since \(K\) is bounded, \(K \subset [-r, r]\) for some \(r > 0\). Since \(K\) is closed (by assumption) and \([-r, r]\) is compact (by the previous theorem), \(K\) is also compact. QED.

For \(\mathbb{R}^n\), replace closed interval by \(n\)-cell.

For example,

  1. Discrete metric on infinite set \(A\): \(A\) is closed and bounded but is not compact.
  2. Let \(\mathcal{C}_b(\mathbb{R})\) is the set of all bounded continuous functions \(f : \mathbb{R} \rightarrow \mathbb{R}\). Let \(d(f, g) = \sup_{x \in \mathbb{R}}|f(x) - g(x)|\).

Properties about Compactness

Theorem. \(K\) is compact if and only if every infinite subset \(E\) of \(K\) has a limit point in \(K\).

Proof.

(\(\Rightarrow\)): If no point in \(K\) is a limit point of \(E\) then each point \(q \in E\) has a neighborhood \(V_q\) containing no other point of \(E\) other than \(q\). \(\{V_q\}\) covers \(E\) with no finite subcover, implying \(K\) is not compact, which is a contradiction.

(\(\Leftarrow\)) proof for \(\mathbb{R}^n\) but is true for all metric spaces: Need to show \(K\) is closed and bounded. Suppose \(K\) is not bounded. Choose \(x_n\) such that \(|x_n| > n\). These have no limit point, which is a contradiction. So \(K\) is bounded. Suppose \(K\) is not closed. There exists a point \(p \notin K\) that is a limit point of \(E\). Choose \(x_n\) such that \(d(x_n, z) < \frac{1}{n}\). \(x_n\) has a limit point at \(p\) and no other. QED.

Corollary (Bolzano-Weierstrass Theorem). Every bounded infinite subset of \(\mathbb{R}^n\) has a limit point in \(\mathbb{R}^n\).

Proof. If subset \(E\) is bounded then \(E\) is in some compact \(k\)-cell. So \(E\) has a limit point in the \(k\)-cell. QED.

A collection of sets has the finite intersection property if any finite subcollection has a non-empty intersection.

Theorem (due to Cantor). Let \(\{K_\alpha\}\) be compact subsets of some metric space \(X\). If \(\{K_\alpha\}\) has the finite intersection property then the intersection of all \(K_\alpha\) is non-empty.

Proof (by contradiction). Let \(U_\alpha = K_\alpha^c\), which is open. Choose an arbitrary \(K\) from \(\{K_\alpha\}\). If \(\bigcap\limits_\alpha K_\alpha = \emptyset\), then \(\{U_\alpha\}\) covers \(K\). Since \(K\) is compact, there exists a finite subcover \(\{U_{\alpha_1}, \dots U_{\alpha_n}\}\) covering \(K\). So, \(K \cap K_{\alpha_1} \cap \dots \cap K_{\alpha_n} = \emptyset\), which constradicts the hypothesis. QED.

Corollary. Let \(\{K_n\}\) be a sequence of compact nested sets. Then \(\bigcap_{n = 1}^\infty K_n\) is non-empty.

Theorem. Any space \(X\) is compact if and only if any collection of closed sets \(\{D_\alpha\}\) that satisfies the finite intersection property has a non-empty intersection (or, if every finite subcollection has non-empty intersection, then \(\bigcap D_\alpha \neq \emptyset\)).

Proof.

(\(\Rightarrow\)): Consider \(\{D_\alpha\}\). These are closed subset of a compact space \(X\), so they are compact. Apply the previous theorem to get the conclusion.

(\(\Leftarrow\), by contrapositive): Assume \(X\) is not compact. Hence, there exists an open cover \(\{U_\alpha\}\) with no finite subcover. Since \(\{U_\alpha\}\) is a cover, every point \(y \in X\) is in \(\bigcup\limits_\alpha U_\alpha\), which implies every point \(y \in X\) is not in \(\bigcap\limits_\alpha U_\alpha\). Hence, \(\bigcap\limits_\alpha U_\alpha = \emptyset\). Consider any subcollection \(\{U_{\alpha_i}\}_{i = 1}^n\). Since \(\{U_{\alpha_i}\}_{i = 1}^n\) is not a subcover of \(X\), there exists \(x \in X \setminus \bigcup\limits_{i = 1}^n U_{\alpha_i}\), which implies \(x \in X \cap \bigcap\limits_{i = 1}^n U_{\alpha_i}\). So \(\bigcap\limits_{i = 1}^n U_{\alpha_i} \neq \emptyset\), as desired. QED.

Perfect Sets

A set is perfect if it is closed and every point is a limit point.

For example,

  1. Any closed interval \([a, b]\) is perfect.
  2. \(\mathbb{R}\) is perfect.

Cantor Sets

Start with \(K_0 = [0, 1]\). Construct \(K_1\) by removing the middle third of \(K_0\), i.e. \(K_1 = \left[0, \frac{1}{3}\right] \cup \left[\frac{2}{3}, 1\right]\). Construct \(K_2\) by removing the middle thirds of the intervals in \(K_1\), i.e. \(K_2 = \left[0, \frac{1}{9}\right] \cup \left[\frac{2}{9}, \frac{3}{9}\right] \cup \left[\frac{6}{9}, \frac{7}{9}\right] \cup \left[\frac{8}{9}, 1\right]\). Continue with the construction.

Each \(K_n\) has \(2^n\) intervals. Each is closed because each is a finite union of closed sets. Each is compact because each is a closed subset of the compact set \([0, 1]\). Finally, all \(K_n\) are nested. Hence, their intersection is non-empty.

Let \(C = \bigcap\limits_{n = 0}^\infty K_n\), called the Cantor set. Notice that

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