# 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,

- 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 \(\left\{W_\frac{5}{10}, W_\frac{6}{10}, W_\frac{7}{10}, W_\frac{8}{10}, W_\frac{8}{10}, W_\frac{9}{10}\right\}\).
- \([0, 1] \in \mathbb{R}\) 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,

- \(\left[\frac{1}{2}, 1\right)\) is not compact (see \(\{V_n\}\)).
- \(\mathbb{Z}\) 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,

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

- Discrete metric on infinite set \(A\): \(A\) is closed and bounded but is not compact.
- 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,

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

- \(C\) is closed because it is the intersection of arbitrary many closed sets.
- \(C\) is perfect.
- \(C\) consists of real numbers whose ternary expansion contains only \(0\)s or \(2\)s, which shows that \(C\) is uncountable (or \(C\) has non-endpoints of \(K_n\)).
- \(C\) has no interior.
- \(C\) is totally disconnected.
- \(C\) has measure \(0\), i.e. for any \(\epsilon > 0\), \(C\) can be covered by intervals of total length less than \(\epsilon\).