The Timpson Lyceum · General Topology (Point Set Topology)

Compactness & Extreme Value Theorem: Last Lecture! – Lecture 22 (Topology)

Overview

This lecture builds toward the central characterization of compactness in Euclidean space and harvests its most useful consequences. It first proves the nested interval theorem and that every closed bounded interval $[a,b]$ is compact, then assembles these into the Heine–Borel theorem: a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. From there it establishes the sequential compactness theorem and the extreme value theorem, finishing by tying together the extreme value and intermediate value theorems to show that continuous functions send closed bounded intervals to closed bounded intervals. The payoff is connecting the abstract open-cover definition of compactness to the familiar "closed and bounded" intuition that makes calculus work.

Key Concepts

Two facts carried over. In a Hausdorff space, every compact set is closed. In any topological space, a closed subset of a compact set is compact. Every metric space is Hausdorff, so both facts apply throughout this lecture.

Compactness (definition). A set $A$ is compact if every open cover of $A$ admits a finite subcover: given any collection of open sets whose union contains $A$, finitely many of them already cover $A$.

Nested intervals. A sequence of closed intervals $[a_n, b_n] \subseteq \mathbb{R}$ is nested if each interval contains the next: $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$. Equivalently, $a_1 \le a_2 \le \cdots \le a_n \le b_n \le \cdots \le b_2 \le b_1.$

• Nested Interval Theorem. For nested closed intervals, $\bigcap_{n=1}^{\infty} [a_n, b_n] \neq \varnothing$. In fact the intersection equals $[A, B]$ where $A = \sup_n a_n$ and $B = \inf_n b_n$; since $A \le B$, this interval is nonempty.

Heine–Borel Theorem (the big theorem). Let $d$ be the standard metric on $\mathbb{R}^n$. A subset $A \subseteq \mathbb{R}^n$ is compact if and only if it is both closed and bounded.

• Compact $\Rightarrow$ closed: immediate, since $\mathbb{R}^n$ is Hausdorff.
• Compact $\Rightarrow$ bounded: the balls $B(a, n)$ for $n \in \mathbb{N}$ cover $A$; a finite subcover sits inside its largest ball, so $A$ is bounded.
• Closed and bounded $\Rightarrow$ compact: a bounded $A$ fits inside a product of closed intervals $B = [-M,M]^n$, which is compact (each $[a,b]$ is compact, and products of compact sets are compact). Then $A$ is a closed subset of the compact set $B$, hence compact.

Sequential Compactness Theorem. Let $(X, d)$ be a metric space and $A$ compact. Every sequence $(x_n)$ in $A$ has a subsequence $(x_{n_m})$ converging to some point of $A$. A subsequence picks infinitely many terms in their original order via a strictly increasing index sequence $n_1 < n_2 < n_3 < \cdots$.

• Key step: there exists a point $a \in A$ such that every open ball $B(a, \varepsilon)$ contains infinitely many terms of the sequence (proved by contradiction—otherwise finitely many balls, each holding finitely many terms, would cover the infinite sequence). Then choosing one term from each $B(a, \tfrac1m)$, later than the last chosen, yields a subsequence converging to $a$.

Extreme Value Theorem (topological form). If $X$ is compact and $f : X \to \mathbb{R}$ is continuous, then $f$ attains a maximum and a minimum: there exist $p, q \in X$ with $f(p) \le f(x) \le f(q)$ for all $x \in X$.

• Supporting lemma: a compact $A \subseteq \mathbb{R}$ contains its supremum and infimum—it has an actual largest and smallest element.
• Proof idea: continuous functions preserve compactness, so $f(X)$ is compact in $\mathbb{R}$, hence closed and bounded, hence contains its max $M$ and min $m$; pull these back through $f$.
• The calculus version (domain $[a,b]$) is the special case where $X$ is a closed bounded interval.

Image of a closed bounded interval. If $f : [a,b] \to \mathbb{R}$ is continuous, then $f([a,b])$ is itself a closed bounded interval $[m, M]$. Compactness gives closed-and-bounded; connectedness of $[a,b]$ plus the Intermediate Value Theorem fills in every value between $m$ and $M$, so the image is one unbroken interval. This is why calculus is done over closed bounded intervals—EVT, IVT, and sequential compactness all behave nicely there.

Worked Example

Claim. The length of the $n$-th interval produced by repeated bisection of $[a,b]$ tends to $0$, which forces the construction to land inside a single open set—the contradiction proving $[a,b]$ is compact.

Suppose an open cover $O$ of $[a,b]$ had no finite subcover. Bisect repeatedly, each time keeping a half $[a_n, b_n]$ that still requires infinitely many sets. The width of the $n$-th interval is the original width halved $n$ times. By the nested interval theorem there is a point $x \in \bigcap_n [a_n, b_n]$; it lies in some $U \in O$, and since $U$ is open there is a radius $\varepsilon$ with $B(x, \varepsilon) \subseteq U$. By the Archimedean property choose $n$ so the width drops below $\varepsilon$:

Since $[a_n, b_n]$ contains $x$ and has width below $\varepsilon$, it lies inside $B(x, \varepsilon) \subseteq U$. But then $[a_n, b_n]$ is covered by the single set $U$—a finite subcover—contradicting that it required infinitely many. Hence $[a,b]$ is compact. $\blacksquare$

Summary

• Nested interval theorem: a decreasing chain of closed intervals always has nonempty intersection, equal to $[\sup a_n,\ \inf b_n]$.
• Every closed bounded interval $[a,b]$ is compact (proved by bisection and contradiction), and finite products of compact sets are compact.
• Heine–Borel: in $\mathbb{R}^n$, compact $\iff$ closed and bounded—the bridge between the open-cover definition and everyday intuition.
• Sequential compactness: in a metric space, every sequence in a compact set has a subsequence converging within the set.
• Extreme value theorem: a continuous real-valued function on a compact space attains its max and min; the classic $[a,b]$ statement is a special case.
• A continuous image of a closed bounded interval is again a closed bounded interval, combining compactness (EVT) with connectedness (IVT).