The Heine-Borel Theorem: Why Compactness Matters in Analysis

Heine-Borel Theorem

A subset $K \subseteq \mathbb{R}^n$ is compact if and only if it is closed and bounded.

• Extreme Value Theorem: A continuous function on a compact set attains its maximum and minimum.
• Uniform Continuity: A continuous function on a compact set is uniformly continuous.
• Bolzano-Weierstrass: Every sequence in a compact set has a convergent subsequence.
• Arzelà-Ascoli: Characterizes compact subsets of $C(X)$.
• Tychonoff's Theorem: Products of compact spaces are compact.
• Brouwer Fixed-Point Theorem: Every continuous self-map of a compact convex set has a fixed point.

Proof that compact implies bounded.

Cover $K$ with the open balls $B(0, n)$ for $n = 1, 2, 3, \ldots$:

$K \subseteq \bigcup_{n=1}^{\infty} B(0, n) = \mathbb{R}^n$

By compactness, finitely many suffice: $K \subseteq B(0, n_1) \cup \cdots \cup B(0, n_N) = B(0, \max n_i)$.

Therefore $K$ is bounded. $\square$

Proof that compact implies closed.

We show $\mathbb{R}^n \setminus K$ is open. Let $x \notin K$. For each $y \in K$, choose disjoint open sets $U_y \ni y$ and $V_y \ni x$ (possible since $\mathbb{R}^n$ is Hausdorff).

The sets $\{U_y\}_{y \in K}$ cover $K$, so by compactness there exist $y_1, \ldots, y_N$ with $K \subseteq U_{y_1} \cup \cdots \cup U_{y_N}$.

Let $V = V_{y_1} \cap \cdots \cap V_{y_N}$. Then $V$ is open, $x \in V$, and $V \cap K = \emptyset$ (since $V \cap U_{y_j} = \emptyset$ for each $j$).

Therefore $x$ is an interior point of $\mathbb{R}^n \setminus K$, so $K$ is closed. $\square$

Proof for $[a, b] \subset \mathbb{R}$ (by bisection)

Let $\{U_\alpha\}$ be an open cover of $[a, b]$. Suppose for contradiction that no finite subcover exists.

Step 1 — Bisect. Divide $[a, b]$ into $[a, m]$ and $[m, b]$ where $m = (a+b)/2$. At least one half has no finite subcover. Call it $[a_1, b_1]$.

Step 2 — Iterate. Repeat: bisect $[a_1, b_1]$ and choose a half with no finite subcover. This gives a nested sequence:

$[a, b] \supset [a_1, b_1] \supset [a_2, b_2] \supset \cdots$

with $b_n - a_n = (b-a)/2^n$ and no $[a_n, b_n]$ finitely coverable.

Step 3 — Nested intervals. By the nested intervals property, there exists a unique point:

$c \in \bigcap_{n=0}^{\infty} [a_n, b_n]$

Step 4 — Contradiction. Since $c \in [a, b]$, there exists $U_\alpha$ containing $c$. Since $U_\alpha$ is open, there exists $\varepsilon > 0$ with $(c - \varepsilon, c + \varepsilon) \subset U_\alpha$.

For large enough $n$, $b_n - a_n < \varepsilon$, so $[a_n, b_n] \subset (c-\varepsilon, c+\varepsilon) \subset U_\alpha$.

But then $\{U_\alpha\}$ is a finite subcover of $[a_n, b_n]$ — contradicting our assumption. $\square$

1. $K$ is compact (every open cover has a finite subcover).
2. $K$ is closed and bounded (Heine-Borel).
3. $K$ is sequentially compact (every sequence has a convergent subsequence in $K$).
4. $K$ is limit point compact (every infinite subset has a limit point in $K$).
5. $K$ is complete and totally bounded (as a metric space).

In infinite-dimensional spaces, the Heine-Borel theorem fails. The closed unit ball in $\ell^2$:

$\overline{B} = \{x \in \ell^2 : \|x\| \leq 1\}$

is closed and bounded but not compact. The sequence $e_1, e_2, e_3, \ldots$ (standard basis vectors) has no convergent subsequence since $\|e_n - e_m\| = \sqrt{2}$ for $n \neq m$.