J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 27: Compact Subspaces of the Real Line

Theorem 27.1

Let \(X\) be a linearly ordered set with the least upper bound property in the order topology. Every closed interval in \(X\) is compact.

Proof

• Let \(a,b \in X\) with \(a < b\). Prove: \([a,b]\) is compact.
• Let \(\mathcal{A}\) be a covering of \([a,b]\) by sets open in \(X\).
• \( \langle 1 \rangle 1 \) For any \(x \in [a,b)\), there exists \(y \in (x,b]\) such that \([x,y]\) can be covered by at most two elements of \(\mathcal{A}\).
• Let \(x \in [a,b)\).
• Pick \(U \in \mathcal{A}\) such that \(x \in U\).
• Pick \(y > x\) such that \([x,y) \subseteq U\).
• Assume w.l.o.g. \(y \leq b\).
• Pick \(V \in \mathcal{A}\) such that \(y \in V\).
• \([x,y]\) is covered by \(U\) and \(V\).
• Let \(C = \{y \in (a,b] \mid [a,y] \text{ can be covered by finitely many elements of } \mathcal{A} \}\).
• Let \(c = \sup C\).
• \(C\) is nonempty.
• Pick \(y \in (a,b]\) such that \([a,y]\) can be covered by at most two elements of \(\mathcal{A}\).
• \(\langle 1 \rangle 1 \)
• \(y \in C\)
• \(c \in C\)
• Pick \(U \in \mathcal{A}\) such that \(c \in U\)
• Pick \(d < c\) such that \((d,c] \subseteq U\)
• Assume w.l.o.g. \(a \leq d\)
• Pick \(e > d\) such that \([a,e]\) can be covered by finitely many elements of \(\mathcal{A}\).
• Pick \(\mathcal{A}_0 \subseteq^{\mathrm{fin}} \mathcal{A}\) that covers \([a,e]\)
• \(\mathcal{A}_0 \cup \{U\}\) covers \([a,c]\)
• \(c = b\)
• Assume for a contradiction \(c < b\)
• Pick \(d > c\) such that \([c,d]\) can be covered by at most two elements of \(\mathcal{A}\).
• \(\langle 1 \rangle 1\)
• \([a,c]\) can be covered by finitely many elements of \(\mathcal{A}\).
• \([a,d]\) can be covered by finitely many elements of \(\mathcal{A}\).
• \(d \in C\)
• This is a contradiction.
• \([a,b]\) can be covered by finitely many elements of \(\mathcal{A}\).
• \(\Box\)

Corollary 27.2

Every closed interval in \(\mathbb{R}\) is compact.

Theorem 27.3

Let \(A \subseteq \mathbb{R}^n\). Then the following are equivalent:

1. \(A\) is compact.
2. \(A\) is closed and bounded in the Euclidean metric.
3. \(A\) is closed and bounded in the square metric.

Proof

• \(1 \Rightarrow 2\) and \(1 \Rightarrow 3\)
• From Theorem 26.3 and Exercise 26.4.
• \(2 \Rightarrow 3\)
• If \(A\) has diameter \(d\) in the Euclidean metric then it has diameter \(
• \(3 \Rightarrow 1\)
• If \(A\) is bounded in the square metric then \(A \subseteq [-K,K]^n\) for some \(K > 0\). If \(A\) is also closed then it is bounded by Theorem 26.2.
• \(\Box\)

Example 1

It follows that the unit spheres \(S^{n-1}\) and closed unit balls \(B^n\) are compact.

Proposition

Let \(X\) be a linearly ordered set in the order topology. If \(X\) is compact, then it has a greatest and a least element.

Proof

• \(X\) has a greatest element.
• Assume for a contradiction \(X\) has no greatest element.
• \(\{(- \infty, x) \mid x \in X \}\) is an open covering of \(X\).
• Pick a finite subcovering \(\{(-\infty, x_1), \ldots, (-\infty, x_n)\}\).
• Let \(a = \max(x_1, \ldots, x_n)\).
• For all \(x \in X\) we have \(x < a\).
• \(a < a\)
• This is a contradiction.
• \(X\) has a least element.
• Similar.
• \(\Box\)

Theorem 27.4 (Extreme Value Theorem)

Let \(X\) be a compact space. Let \(Y\) be a linearly ordered set in the order topology. Let \(f : X \rightarrow Y\) be continuous. Then there exist \(c,d \in X\) such that \(\forall x \in X. f(c) \leq f(x) \leq f(d)\).

Proof

Since \(f(X)\) is compact (Theorem 26.5) and therefore has a greatest and a least element.

Definition

Let \(X\) be a metric space. Let \(A\) be a nonempty subset of \(X\) and \(x \in X\). The distance from \(x\) to \(A\) is

\[ d(x,A) = \inf_{a \in A} d(x,a) \enspace . \]

Definition

Let \(X\) be a metric space. Let \(\mathcal{A}\) be an open covering of \(X\). A Lebesgue number for \(\mathcal{A}\) is a real number \(\delta > 0\) such that every set of diameter \(< \delta\) is included in some element of \(\mathcal{A}\).

Proposition

Let \(X\) be a metric space. Let \(A\) be a nonempty subset of \(X\). The function \(d(-,A) : X \rightarrow \mathbb{R}\) is continuous.

Proof

• For all \(x,y \in A\) we have \(|d(x,A) - d(y,A)| \leq d(x,y)\).
• Let \(x,y \in A\).
• For all \(a \in A\) we have \(d(x,A) \leq d(x,y) + d(y,a)\)
• \(d(x,A) \leq d(x,a) \leq d(x,y) + d(y,a)\)
• \(d(x,A) - d(x,y) \leq d(y,A)\)
• \(d(x,A) - d(y,A) \leq d(x,y)\)
• \(d(y,A) - d(x,A) \leq d(x,y)\)
• Let \(x \in X\) and \(\epsilon > 0\).
• For all \(y \in X\), if \(d(x,y) < \epsilon\) then \(|d(x,A) - d(y,A)| < \epsilon\).

Lemma 27.5 (The Lebesgue Number Lemma)

Every open covering of a compact metric space has a Lebesgue number.

Proof

• Let \(X\) be a compact metric space.
• Let \(\mathcal{A}\) be an open covering of \(X\).
• Assume w.l.o.g. \(X \notin \mathcal{A}\).
• Pick a finite subcovering \(\{A_1, \ldots, A_n\}\).
• For \(i = 1, \ldots, n\), let \(C_i = X - A_i\).
• Define \(f : X \rightarrow \mathbb{R}\) by
\[ f(x) = \frac{1}{n} \sum_{i=1}^n d(x, C_i) \enspace . \]
• For all \(x \in X\) we have \(f(x) > 0\).
• Let \(x \in X\)
• Pick \(i\) such that \(x \in A_i\)
• Pick \(\epsilon > 0\) such that \(B(x,\epsilon) \subseteq A_i\)
• \(d(x,C_i) \geq \epsilon\)
• \(f(x) \geq \epsilon / n\)
• Let \(\delta\) be the minimum value of \(f\).
• Extreme Value Theorem
• Let \(B\) be a nonempty subset of \(X\) of diameter \(< \delta\).
• Pick \(x_0 \in B\)
• Pick \(m\) such that \(d(x_0, C_m) = \max(d(x_0, C_1), \ldots, d(x_0, C_n)\)
• \(B \subseteq A_m\)
• Let \(y \in B\)
• \(d(x_0,y) < d(x_0,C_m)\)
\[ \begin{align} d(x_0,y) & \leq \operatorname{diam} B \\ & < \delta \\ & \leq f(x_0) \\ & = 1/n (d(x_0,C_1) + \cdots + d(x_0,C_n)) \\ & \leq 1/n(d(x_0,C_m) + \cdots + d(x_0,C_m)) \\ & = d(x_0,C_m) \end{align} \]
• \(y \notin C_m\)
• \(y \in A_m\)
• \(\Box\)

Definition

Let \(X\) and \(Y\) be metric spaces and \(f : X \rightarrow Y\). Then \(f\) is uniformly continuous iff, for every \(\epsilon > 0\), there exists \(\delta > 0\) such that, for all \(x,y \in X\), if \(d(x,y) < \delta\) then \(d(f(x),f(y)) < \epsilon\).

Theorem 27.6 (Uniform Continuity Theorem)

Every continuous function from a compact metric space to a metric space is uniformly continuous.

Proof

• Let \(X\) be a compact metric space and \(Y\) a metric space. Let \(f : X \rightarrow Y\) be continuous.
• Let \(\epsilon > 0\).
• Let \(\mathcal{A} = \{f^{-1}(B(y,\epsilon/2)) \mid y \in Y \}\)
• Let \(\delta\) be a Lebesgue number for \(\mathcal{A}\).
• Let \(x,y \in X\). Assume \(d(x,y) < \delta\).
• Pick \(z \in Y\) such that \(\{x,y\} \subseteq f^{-1}(B(z,\epsilon/2))\)
• \(d(x,y) < \epsilon\)
• \(\Box\)

Definition

Let \(X\) be a topological space and \(x \in X\). Then \(x\) is an isolated point iff \(\{x\}\) is open.

Theorem 27.7

Every nonempty compact Hausdorff space with no isolated points is uncountable.

Proof

• Let \(X\) be a nonempty compact Hausdorff space with no isolated points.
• For any nonempty open set \(U\) in \(X\) and any point \(x \in X\), there exists a nonempty open set \(V \subseteq U\) such that \(x \notin \overline{V}\).
• Let \(U\) be a nonempty open set and \(x \in X\).
• Pick \(y \in U\) such that \(y \neq x\).
• We cannot have \(U = \{x\)\) because \(X\) has no isolated points.
• Pick disjoint open sets \(W_1\) and \(W_2\) with \(x \in W_1\) and \(y \in W_2\).
• Let \(V = W_2 \cap U\)
• \(x \notin \overline{V}\)
• Since \(\overline{V} \subseteq X - W_1\).
• Let \(f : \mathbb{Z}_+ \rightarrow X\). Prove: \(f\) is not surjective.
• Pick a sequence of open sets \((V_n)\) such that \(V_{n+1} \subseteq V_n\) for all \(n\) and \(f(n) \notin \overline{V_n}\) for all \(n\).
• Pick \(x \in \bigcap_n \overline{V_n}\)
• Theorem 26.9
• \(f(n) \neq x\) for all \(n\).
• \(\Box\)

Corollary 27.8

Every closed interval in \(\mathbb{R}\) with more than one point is uncountable.

Exercises

Exercise 1

Prove that if \(X\) is a linearly ordered set in which every closed interval in compact, then \(X\) has the least upper bound property.

Solution

• Let \(A\) be a nonempty subset of \(X\) bounded above.
• Pick \(a \in A\) and pick an upper bound \(u\) of \(A\).
• Assume for a contradiction \(A\) has no least upper bound.
• \(\{(-\infty,a) \mid a \in A\} \cup \{(u, +\infty) \mid u \text{ is an upper bound for } A \}\) is a covering of \([a,u]\) by open sets in \(X\).
• Let \(x \in [a,u]\)
• Assume \(x \notin (-\infty, a)\) for any \(a \in A\)
• \(x\) is an upper bound for \(A\)
• Pick an upper bound \(u'\) for \(A\) such that \(u' < x\)
• \(x \in (u',+\infty)\)
• Pick a finite subcovering \(\{(-\infty, a_1), \ldots, (-\infty,a_m), (u_1, +\infty), \ldots, (u_n, +\infty)\}\).
• Let \(a = \max(a_1, \ldots, a_m)\) and \(s = \min(u_1, \ldots, u_n)\).
• \(s \notin (-\infty, a_i)\) for any \(i\)
• \(s \notin (u_i, +\infty)\) for any \(i\)
• This is a contradiction.
• \(\Box\)

Exercise 2

Let \((X,d)\) be a metric space. Let \(A \subseteq X\) be nonempty.

(a)

Show that \(d(x,A)=0\) if and only if \(x \in \overline{A}\).

Solution

The following are equivalent:

• \(d(x,A) = 0\)
• For all \(\epsilon > 0\), there exists \(a \in A\) such that \(d(x,a) < \epsilon\)
• For all \(\epsilon > 0\), we have \(B(x,\epsilon)\) intersects \(A\)
• \(x \in \overline{A}\)
• \(\Box\)

(b)

Show that if \(A\) is compact then, for all \(x \in X\), there exists \(a \in A\) such that \(d(x,A) = d(x,a)\).

Solution

This follows from the Extreme Value Theorem and the fact that \(d(x,-)\) is continuous.

(c)

Define the \(\epsilon\)-neighbourhood of \(A\) in \(X\) to be the set

\[ U(A,\epsilon) = \{ x \in X \mid d(x,A) < \epsilon \} \enspace . \]

Show that \(U(A,\epsilon)\) equals the union of the open balls \(B_d(a, \epsilon)\) for \(a \in A\).

Solution

\[ \begin{align} x \in U(A, \epsilon) & \Leftrightarrow d(x,A) < \epsilon \\ & \Leftrightarrow \exists a \in A. d(x,a) < \epsilon \\ & \Leftrightarrow \exists a \in A. x \in B(a, \epsilon) \end{align} \]

(d)

Assume that \(A\) is compact; let \(U\) be an open set containing \(A\). Show that some \(\epsilon\)-neighbourhood of \(A\) is contained in \(U\).

Solution

• For all \(a \in A\), there exists \(\epsilon > 0\) such that \(B(a,\epsilon) \subseteq U\).
• \(\{B(a, \epsilon) \mid a \in A, \epsilon > 0, B(a, \epsilon) \subseteq U\}\) is a covering of \(A\) by open sets in \(X\).
• Pick a Lebesgue number \(\delta\) for this covering.
• For all \(a \in A\) we have \(B(a, \delta / 2\) \subseteq U\)
• Let \(x \in B(a, \delta / 2)\).
• Pick \(a'\) and \(\epsilon\) such that \(B(a, \delta / 2) \subseteq B(a', \epsilon) \cap A\) and \(B(a', \epsilon) \subseteq U\).
• \(x \in U\)
• \(U(A, \delta / 2) \subseteq U\)
• \(\Box\)

(e)

Show that the result in (d) need not hold if \(A\) is closed but not compact.

Solution

Let \(A = \{(x, 1/x) \mid 0 < x \leq 1\}\) and \(U = (0,2) \times \mathbb{R}\). Then \(A \subseteq U\). However, there is no \(\epsilon > 0\) such that \(B(A, \epsilon) \subseteq U\). For let \(\epsilon > 0\). Then \((2/\epsilon, \epsilon/2\) \in A\). So \((2/\epsilon, -\epsilon/4) \in U(A, \epsilon) - U\).

Exercise 3

Recall that \(\mathbb{R}_K\) denotes \(\mathbb{R}\) in the \(K\)-topology.

(a)

Show that \([0,1]\) is not compact as a subspace of \(\mathbb{R}_K\).

Solution

We have \(\{[0,1]-K, (1/2,1], (1/3,1),(1/4,1/2), \ldots,\}\) is an open covering of \([0,1]\) but has no finite subcovering.

(b)

Show that \(\mathbb{R}_K\) is connected.

Solution

• Assume for a contradiction \(\mathbb{R}_K = U \cup V\) is a separation of \(\mathbb{R}_K\).
• Assume w.l.o.g. \((-\infty, 0) \subseteq U\).
• Lemma 23.2
• We have \((0, +\infty) \subseteq V\).
• Lemma 23.2, and we cannot have \((0, +\infty) \subseteq U\) because then \(V = \{0\}\) which is not open in \(\mathbb{R}_K\).
• We have \(0 \in U\) or \(0 \in V\)
• We have \(U = (-\infty,0]\) or \(V = [0, +\infty)\).
• This is a contradiction, as neither of these sets is open in \(\mathbb{R}_K\).
• \(\Box\)

(c)

Show that \(\mathbb{R}_K\) is not path connected.

Solution

• Assume for a contradiction \(p : [0,1] \rightarrow \mathbb{R}_K\) is a path from 0 to 1.
• \(p([0,1])\) is connected.
• \([0,1] \subseteq p([0,1])\)
• \(p([0,1])\) is compact in \(\mathbb{R}_K\).
• \([0,1]\) is compact in \(\mathbb{R}_K\).
• This contradicts part (a).
• \(\Box\)

Exercise 4

Show that a connected metric space having more than one point is uncountable.

Solution

• Let \(X\) be a connected metric space with more than one point.
• Pick \(a \in X\)
• \(d(a,-) : X \rightarrow \mathbb{R}\) is continuous.
• \(\{d(a,x) \mid x \in X\}\) is a connected subspace of \(\mathbb{R}\) with more than one point.
• \(\{d(a,x) \mid x \in X\}\) is uncountable.
• \(X\) is uncountable.
• \(\Box\)

Exercise 5

Let \(X\) be a compact Hausdorff space. Let \((A_n)\) be a sequence of closed sets in \(X\). Show that if each \(A_n\) has empty interior, then the union \(\bigcup_n A_n\) has empty interior.

Solution

• \( \langle 1 \rangle 1 \) For any closed set \(A\) in \(X\) and open set \(U\) with \(U \nsubseteq A\), there exists a nonempty open set \(V\) such that \(\overline{V} \subseteq U - A\).
• Let \(A\) be any closed set and \(U\) an open set with \(U \nsubseteq A\).
• Pick \(x \in U - A\).
• Pick disjoint open sets \(W\) and \(V\) with \(A \cup (X - U) \subseteq W\) and \(x \in V\).
• Lemma 26.4.
• \(\overline{V} \subseteq U - A\)
• \(\overline{V} \subseteq X - W \subseteq U - A\)
• Let \(U_1\) be any nonempty open set. Prove: \(U_1 \nsubseteq \bigcup_n A_n\).
• Extend \(U_1\) to a sequence of nonempty open sets \((U_n)\) such that \(\overline{U_{n+1}} \subseteq U_n - A_n\) for all \(n \in \mathbb{Z}_+\).
• Assume we have chosen \(U_1, \ldots, U_n\) with \(\overline{U_{k+1}} \subseteq U_k - A_k\) for \(k < n\).
• \(U_n \nsubseteq A_n\)
• Since \(A_n\) has empty interior.
• Pick a nonempty open set \(U_{n+1}\) such that \(\overline{U_{n+1}} \subseteq U_n - A_n\).
• \(\langle 1 \rangle 1\)
• Pick \(a \in \bigcap_n \overline{U_n}\).
• Theorem 26.9.
• \(a \in U_1\)
• Since \(a \in \overline{U_2} \subseteq U_1\).
• \(a \notin A_n\) for any \(n\).
• Since \(a \in \overline{U_{n+1}} \subseteq U_n - A_n\).
• \(\Box\)

Exercise 6

Let \(A_0\) be the closed interval \([0,1]\) in \(\mathbb{R}\). Let \(A_1\) be the set obtained from \(A_0\) by deleting its "middle third" \((1/3, 2/3)\). Let \(A_2\) be the set obtained from \(A_1\) by deleting its "middle thirds" \((1/9,2/9)\) and \((7/9,8/9)\). In general, define \(A_n\) by the equation

\[ A_n = A_{n-1} - \bigcup_{k=0}^\infty \left( \frac{1 + 3k}{3^n}, \frac{2+3k}{3^n} \right) \enspace . \]

The intersection

\[ C = \bigcap_{n \in \mathbb{Z}_+} A_n \]

is called the Cantor set; it is a subspace of \([0,1]\).

(a)

Show that \(C\) is totally disconnected.

Solution

• Let \(A \subseteq C\) have more than one point. Prove: \(A\) is disconnected.
• Pick \(x,y \in A\) with \(x \neq y\).
• Pick \(n\) such that \(1/3^n < |y-x|\)
• \(x\) and \(y\) are in different closed intervals in \(A_n\).
• Since each closed interval in \(A_n\) has length \(1/3^n\).
• Pick \(z\) between \(x\) and \(y\) such that \(z \notin C\).
• \((- \infty, z)\) and \((z, + \infty)\) form a separation of \(A\).
• \(\Box\)

(b)

Show that \(C\) is compact.

Solution

It is a closed subspace of the compact set \([0,1]\).

(c)

Show that each \(A_n\) is a union of finitely many disjoint closed intervals of length \(1/3^n\); and show that the end points of these intervals lie in \(C\).

Solution

Immediate from the definition.

(d)

Show that \(C\) has no isolated points.

Solution

• Let \(a \in C\).
• Let \((I_n)\) be the sequence of intervals such that \(I_n\) is one of the intervals that make up \(A_n\) and \(a \in I_n\).
• Let \((b_n)\) be a sequence of endpoints of \(I_n\) such that \(b_n \neq a\).
• \(b_n \rightarrow a\) as \(n \rightarrow \infty\)
• \(a\) is a limit point of \(C - \{a\}\).
• Every open set in \(C\) containing \(a\) contains other points.
• \(\Box\)

(e)

Conclude that \(C\) is uncountable.

Solution

From Theorem 27.7.