Topological Boundedness

We have previously discussed how $[a, b]$ is a closed set and formulated the general version of closed sets in topology, demonstrating several useful properties that they satisfy. For boundedness, we say that $[a, b]$ is bounded because there is an absolute value function such that $|x| \leq M$ . But topologically, we don’t have such a norm (or rather, a metric), at least not in the general setting. How then do we describe the boundedness of $[a ,b]$ ?

We cheat—we combine the closed AND boundedness of the set into compactness.

Theorem 1. Let $\mathcal V := \{ V_\alpha \}$ be a collection of open sets in $\mathbb R$ such that $\bigcup_\alpha V_\alpha \supseteq [a, b]$ . (We call $\mathcal V$ an open cover for $[a, b]$ .) Then there exists a finite sub-collection $\{V_1,\dots,V_n\} \subseteq \mathcal V$ of open sets such that $\bigcup_{i=1}^n V_i \supseteq [a, b]$ .

Proof. Fix any open cover $\{ V_\alpha \}$ of $[a, b]$ . Define the subset $K \subseteq [a, b]$ to be the set of $x \in [a, b]$ such that $[a , x]$ can be covered by finitely many $V_i$ . We will verify that the three hypotheses of “real induction” hold, so that $K = [a, b]$ , as required.

For the first hypothesis, $[a,a] = \{a\} \subseteq V_m$ for some open set $V_m$ , so that $a \in K$ .

For the second hypothesis, suppose $[a, x] \subseteq K$ . Find a finite subcover $\{V_1,\dots, V_m\}$ of $[a, x]$ . In particular, suppose without loss of generality that $x \in V_m$ . Since $V_m$ is open, there exists $\delta > 0$ such that $[x,x+\delta] \subseteq V_m$ . This means for any $y \in [x, x+\delta]$ , $[a,y]$ is covered by $\{V_1,\dots, V_m\}$ . Therefore, $[x, x+\delta] \subseteq K$ .

For the final hypothesis, suppose $[a,x) \subseteq K$ . Suppose $x \in V_0$ . Since $V_0$ is open, there exists $\delta > 0$ such that $(x-\delta, x+\delta) \subseteq V_0$ . Since $x-\delta/2 \in [a,x) \subseteq K$ , find a finite sub-collection $\{V_1,\dots, V_m\}$ that covers $[a,x-\delta/2]$ . Then the finite sub-collection $\{V_0, V_1,\dots, V_m\}$ covers $[a, x] = [a,x-\delta/2] \cup [x-\delta/2,x]$ , which implies $x \in K$ .

By the principle of “real induction”, $K = [a, b]$ , as required.

Roughly speaking, this theorem states that no matter how we cover $[a, b]$ using open sets, we can always extract a finite sub-collection of them to do the job. In this regard, compact sets share many of the properties of finite sets, except that they aren’t actually finite at all.

Let $K$ be any topological space.

Definition 1. We say that $C \subseteq K$ is compact if any (possibly infinite) open cover $\{V_\alpha\}$ of $C$ admits a finite sub-cover $\{V_1,\dots,V_n\}$ . In particular, closed and bounded sets of the form $[a, b]$ are compact in $\mathbb R$ . In particular, if $K$ is compact, then we call $K$ a compact space.

Compactness is probably one of the most useful properties in all of mathematics, since it allows us to impose finite-ish arguments onto infinite-ish sets by passing into the finite sub-covers. In particular, we will show in the context of $\mathbb R$ that compact sets are closed and bounded.

Lemma 1. Suppose $K$ is compact. If $C \subseteq K$ is closed, then it is compact.

Proof. Let $\{V_\alpha\}$ be any open cover of $C$ . Since $C$ is closed, $K \backslash C$ is open. Hence, $\{V_\alpha\} \cup \{K \backslash C\}$ forms an open cover for $K$ . Since $K$ is compact, it admits a finite sub-cover $\{V_1,\dots,V_n\}$ of $K$ . Therefore, $\{V_1,\dots,V_n\} \backslash \{ K \backslash C \}$ forms a finite sub-cover for $C$ , and $C$ is compact.

Theorem 2. If $C \subseteq \mathbb R$ is closed and bounded, then it is compact.

Proof. Since $C$ is bounded, there exists $r > 0$ such that $C \subseteq [-r , r]$ . Since $[-r, r]$ is compact by Theorem 1 and $C$ is closed, by Lemma 1, $C$ is compact.

Our final goal is to proceed in the opposite direction: any compact set in $\mathbb R$ must be closed and bounded.

Lemma 2. Suppose $K$ is Hausdorff. If $C \subseteq K$ is compact, then it is closed.

Proof. We need to prove that $K \backslash C$ is open. Fix $x \in K \backslash C$ . Since $K$ is Hausdorff, for any $y \in C$ , there exists disjoint neighbourhoods $U_y, V_y$ of $x, y$ respectively. Then $\{V_y\}$ forms an open cover for $C$ . Since $C$ is compact, there exists a finite subset $\{y_1,\dots,y_n\}$ such that $\{V_{y_i}\}$ covers $C$ . We note that the set $U := \bigcap_{i=1}^n U_{y_i}$ is still a neighbourhood of $x$ that is disjoint with $V := \bigcup_{i=1}^n V_{y_i}$ . Therefore, $U \subseteq K \backslash V \subseteq K \backslash C$ . Therefore, $K \backslash C$ is open, as required.

Theorem 3. If $C \subseteq \mathbb R$ is compact, it is closed and bounded.

Proof. By Lemma 2, since $\mathbb R$ is a metric space and thus Hausdorff, $C$ is closed. Fix $\epsilon = 1$ . For each $x \in C$ , the open set $V_x := (x-\epsilon, x+\epsilon)$ covers $\{x\}$ , so that the collection $\{ V_x \}$ forms an open cover for $C$ . By compactness, there exist finitely many $\{x_1,\dots,x_n\}$ (assume they are arranged in ascending order) such that $\{V_{x_i}\}$ covers $C$ . Hence, for any $x \in C$ , $x_1 - \epsilon < x < x_n + \epsilon$ . Finding $M > \max \{|x_1 - \epsilon|, |x_n + \epsilon|\}$ yields $|x| \leq M$ so that $C$ is bounded.

We haven’t done a lot to generalise compactness yet, since we want to ensure that our ideas of compactness really do correspond with closed and bounded sets, at least in the context of $\mathbb R$ . We’ll do more of that in future posts. For now, let’s state and prove and contextualise our topological reformulation of the extreme value theorem.

Theorem 4 (Extreme Value Theorem). Let $L$ be a topological space and $f : K \to L$ be a continuous map. If $C \subseteq K$ is compact, so is $f(C)$ . In particular, if $f : [a, b] \to \mathbb R$ is continuous, then $f([a, b])$ is closed and bounded.

Proof. For any open cover $\{V_\alpha\}$ of $f(C)$ , since $f$ is continuous $\{ f^{-1}(V_\alpha) \}$ is an open cover of $C$ . Since $C$ is compact, there exists a finite sub-cover $\{f^{-1}(V_1),\dots,f^{-1}(V_n)\}$ of $C$ . Hence, $\{V_1,\dots,V_n\}$ forms a finite sub-cover for $f(C)$ , and $f(C)$ is compact.

There is something slightly disconcerting about this result. Doesn’t the original result state that we can find $c_1,c_2 \in [a, b]$ such that $f([a ,b]) = [f(c_1), f(c_2)]$ ? But we only have $f([a, b])$ being compact, and not an interval. It turns out that we need to investigate another property about the set $[a, b]$ , namely that it is connected, in order to make the interval-based conclusion.

This connectedness property also turns out to be responsible for the intermediate value theorem in real analysis. In light of this unanswered question, we will pivot to connectedness in the next post, establish the intermediate value theorem and extreme value theorem therein, before returning to explore compactness.

—Joel Kindaik, 31 Mar 25, 1354H

KindiakMath

Topological Boundedness

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Topological Boundedness

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