Open, Closed and Compact Sets (the Heine-Borel Theorem)

Compactness is the single most useful finiteness property in analysis: it lets an argument that only works for finitely many objects extend to infinitely many. Continuous functions attain their maximum on compact sets, uniform continuity holds there automatically, and whole classes of existence theorems (the Extreme Value Theorem, among others) reduce to this one idea. This chapter defines compactness via open covers — the definition that generalises beyond $\mathbb{R}^n$ — proves its equivalence with sequential compactness in metric spaces, and proves the Heine-Borel theorem characterising compact subsets of $\mathbb{R}$ (and, by the same coordinatewise pattern, $\mathbb{R}^n$) concretely as the closed and bounded ones.

Have to hand: open and closed sets, limit points and sequences in a metric space (Chapter 2), and the Archimedean property and nested interval property (Chapter 1) — the bisection argument of Section 3 reuses the latter directly.

Why open covers, and not just "closed and bounded"? "Closed and bounded" is a consequence of compactness in $\mathbb{R}^n$, true by the Heine-Borel theorem — but it is specific to $\mathbb{R}^n$. In general metric (and later topological) spaces, closed and bounded sets can fail to be compact. The open-cover definition is what actually generalises, which is why it is taken as primary.

1. Open Covers and Compactness

Definition. An open cover of $K\subseteq X$ is a collection $\{U_\alpha\}_{\alpha\in A}$ of open sets with $K\subseteq\bigcup_{\alpha\in A}U_\alpha$. A subcover is a subcollection that still covers $K$.

$K$ is compact if every open cover of $K$ has a finite subcover.

Worked example — a non-compact set via a bad cover. Let $K=(0,1)$ and consider the open cover $U_n = (1/n, 1)$, $n\in\mathbb{N}$. Every $x\in(0,1)$ lies in some $U_n$ (take $n$ with $1/n<x$, by the Archimedean property), so this is indeed a cover. But any finite subcollection $U_{n_1},\ldots,U_{n_k}$ equals $U_N$ where $N=\max(n_1,\ldots,n_k)$ (the largest index gives the largest interval, since $U_n$ grows as $n$ grows), and $U_N=(1/N,1)$ misses $(0,1/N]$. No finite subcover exists, so this particular cover already shows $(0,1)$ is not compact.

Key distinction. To prove a set is not compact, exhibiting one open cover with no finite subcover suffices (as above). To prove a set is compact, you must handle every possible open cover — which is why direct open-cover arguments are hard, and why the sequential and Heine-Borel characterisations below are so valuable.

2. Sequential Compactness

Definition. $K\subseteq X$ is sequentially compact if every sequence in $K$ has a subsequence converging to a point of $K$.

A subsequence of $(x_n)$ is a sequence $(x_{n_k})_{k=1}^{\infty}$ where $n_1<n_2<n_3<\cdots$ is a strictly increasing sequence of indices; it keeps infinitely many of the original terms, in order, discarding the rest.

Theorem. In a metric space, $K$ is compact if and only if $K$ is sequentially compact.

Proof sketch (compact $\Rightarrow$ sequentially compact). Suppose $K$ is compact and $(x_n)$ is a sequence in $K$ with no subsequence converging in $K$. Then for every $y\in K$, some ball $B(y,r_y)$ contains $x_n$ for only finitely many $n$ (otherwise a subsequence could be extracted converging to $y$). The collection $\{B(y,r_y)\}_{y\in K}$ is an open cover of $K$; by compactness finitely many balls $B(y_1,r_{y_1}),\ldots,B(y_m,r_{y_m})$ cover $K$. But each contains only finitely many terms of $(x_n)$, so their union contains only finitely many terms in total — contradicting that all $x_n\in K$ lie in this union.

Proof (sequentially compact $\Rightarrow$ compact). This direction needs two lemmas; neither is the bisection argument of Section 3 (that argument only produces Bolzano–Weierstrass in $\mathbb{R}$ and does not by itself yield a finite subcover in a general metric space).

Lemma A (totally bounded). If $K$ is sequentially compact, then for every $\varepsilon>0$, $K$ is covered by finitely many balls of radius $\varepsilon$. Proof. If not, some $\varepsilon>0$ admits no finite cover by $\varepsilon$-balls. Pick $x_1\in K$ arbitrarily; having chosen $x_1,\ldots,x_n$, the balls $B(x_1,\varepsilon),\ldots,B(x_n,\varepsilon)$ do not cover $K$ (else they would be a finite subcover), so pick $x_{n+1}\in K$ outside all of them. The resulting sequence satisfies $d(x_i,x_j)\ge\varepsilon$ for all $i\ne j$, so no subsequence is Cauchy, hence none converges — contradicting sequential compactness.

Lemma B (Lebesgue number). If $K$ is sequentially compact and $\{U_\alpha\}$ is an open cover of $K$, there exists $\delta>0$ such that every subset of $K$ of diameter less than $\delta$ lies entirely inside some single $U_\alpha$. Proof. If not, for each $n\in\mathbb{N}$ there is $S_n\subseteq K$ with $\mathrm{diam}(S_n)<1/n$ contained in no single $U_\alpha$; pick $x_n\in S_n$. By sequential compactness, some subsequence $x_{n_k}\to x\in K$. Since $\{U_\alpha\}$ covers $K$, $x\in U_\alpha$ for some $\alpha$, and openness gives $B(x,r)\subseteq U_\alpha$ for some $r>0$. For $k$ large enough that $d(x_{n_k},x)<r/2$ and $1/n_k<r/2$: any $y\in S_{n_k}$ satisfies $d(y,x)\le d(y,x_{n_k})+d(x_{n_k},x)<1/n_k+r/2<r$, so $S_{n_k}\subseteq B(x,r)\subseteq U_\alpha$ — contradicting that $S_{n_k}$ lies in no single $U_\alpha$.

Main argument. Let $\{U_\alpha\}$ be any open cover of $K$, and let $\delta>0$ be a Lebesgue number for it (Lemma B). By Lemma A with $\varepsilon=\delta/3$, $K$ is covered by finitely many balls $B(x_1,\delta/3),\ldots,B(x_m,\delta/3)$. Each set $B(x_i,\delta/3)\cap K$ has diameter at most $2\delta/3<\delta$, so by the defining property of $\delta$ it lies inside some single $U_{\alpha_i}$. Then $U_{\alpha_1},\ldots,U_{\alpha_m}$ is a finite subcover of $K$. $\blacksquare$

This equivalence is why, in practice, compactness in $\mathbb{R}^n$ is almost always checked via sequences rather than open covers directly.

3. The Bolzano–Weierstrass Theorem

Theorem (Bolzano–Weierstrass). Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

Proof. Let $(x_n)$ be bounded, so $x_n\in[a,b]$ for all $n$, some $a<b$. Bisect $[a,b]$ into two closed halves; at least one half contains $x_n$ for infinitely many $n$ (if both contained only finitely many, the whole interval would too, contradicting that every $n$ appears). Call this half $I_1=[a_1,b_1]$, and pick an index $n_1$ with $x_{n_1}\in I_1$. Bisect $I_1$ again; again one half contains infinitely many of the remaining terms. Call it $I_2\subseteq I_1$, and pick $n_2>n_1$ with $x_{n_2}\in I_2$ (possible since $I_2$ contains infinitely many terms, so some index beyond $n_1$ works). Repeating indefinitely gives nested closed intervals $I_1\supseteq I_2\supseteq\cdots$ with $b_k-a_k = (b-a)/2^k \to 0$, and indices $n_1<n_2<\cdots$ with $x_{n_k}\in I_k$.

By the Nested Interval Property (Chapter 1), $\bigcap_k I_k = \{L\}$ for a single point $L$. Since $x_{n_k},L\in I_k$ and $I_k$ has length $(b-a)/2^k$, $|x_{n_k}-L|\le (b-a)/2^k \to 0$ by the Archimedean property, so $x_{n_k}\to L$. $\blacksquare$

This is the engine behind the Heine-Borel theorem and reappears constantly: whenever a sequence is merely known to be bounded, Bolzano–Weierstrass upgrades that to "has a convergent subsequence," which is often exactly what an existence proof needs.

4. The Heine-Borel Theorem

Theorem (Heine-Borel, in $\mathbb{R}$). $K\subseteq\mathbb{R}$ is compact if and only if $K$ is closed and bounded.

Proof. ($\Rightarrow$) Suppose $K$ is compact. Bounded: the cover $\{(-n,n)\}_{n\in\mathbb{N}}$ of $\mathbb{R}$ (hence of $K$) has a finite subcover, which (being a finite collection of these nested intervals) equals some single $(-N,N)$; so $K\subseteq(-N,N)$ is bounded. Closed: suppose $x\notin K$. For each $y\in K$, let $r_y = |y-x|/2>0$; the balls $B(y,r_y)$ cover $K$ (each $y$ lies in its own ball), so finitely many $B(y_1,r_{y_1}),\ldots,B(y_m,r_{y_m})$ cover $K$. Let $r=\min(r_{y_1},\ldots,r_{y_m})>0$; one checks $B(x,r)$ is disjoint from each $B(y_i,r_{y_i})$ (since $d(x,y_i) = 2r_{y_i}\ge r_{y_i}+r$, so points within $r$ of $x$ cannot be within $r_{y_i}$ of $y_i$), hence disjoint from $K$. So $X\setminus K$ is open, i.e. $K$ is closed.

($\Leftarrow$) Suppose $K$ is closed and bounded. By Section 2, it suffices to show $K$ is sequentially compact. Let $(x_n)$ be a sequence in $K$. Since $K$ is bounded, Bolzano–Weierstrass gives a subsequence $x_{n_k}\to L$ for some $L\in\mathbb{R}$. Since $K$ is closed and $(x_{n_k})$ is a sequence in $K$ converging to $L$, the sequential characterisation of closed sets (Chapter 2) gives $L\in K$. So every sequence in $K$ has a subsequence converging to a point of $K$: $K$ is sequentially compact, hence compact. $\blacksquare$

Scope warning. Heine-Borel as stated is specific to $\mathbb{R}$ (and, with the same proof pattern applied coordinatewise, to $\mathbb{R}^n$). In a general metric space, closed and bounded does not imply compact — this is explored in the exercises, and is the reason the open-cover definition, not "closed and bounded," is taken as the primary meaning of compactness.

Worked example. $K=[0,1]\cup\{3\}$ is compact: it is closed (finite union of the closed set $[0,1]$ and the closed singleton $\{3\}$) and bounded (contained in $[0,3]$), so Heine-Borel applies. By contrast $(0,1)$ is bounded but not closed (missing limit points $0,1$), consistent with the explicit failure of compactness shown in Section 1.

5. Compactness, Closedness and Subsets

Theorem. (i) A compact subset of a metric space is closed. (ii) A closed subset of a compact set is compact.

Proof sketch. (i) is the "closed" half of the Heine-Borel proof above, which did not in fact use boundedness of $X$ itself — it only used compactness of $K$, so it holds in any metric space. (ii) Let $F\subseteq K$ be closed with $K$ compact, and let $\{U_\alpha\}$ be an open cover of $F$. Adding the open set $X\setminus F$ to this collection covers all of $K$ (every point of $K$ is either in $F$, covered by some $U_\alpha$, or outside $F$, covered by $X\setminus F$). Compactness of $K$ gives a finite subcover of $K$, hence of $F$; discard $X\setminus F$ from it if present (it contributes nothing to covering $F$), leaving a finite subcover of $F$ drawn from the $U_\alpha$. $\blacksquare$

Common Exam Pitfalls

• Treating "closed and bounded $\Rightarrow$ compact" as universal. It is a theorem specific to $\mathbb{R}^n$ (Heine-Borel); do not quote it in a general metric space without justification.
• Proving compactness by checking only one open cover. Compactness requires every open cover to have a finite subcover; checking one well-chosen cover only ever disproves compactness, never proves it.
• Confusing "finite subcover" with "finitely many points." A finite subcover is a finite subcollection of the given open sets that still covers $K$ — not a statement that $K$ itself is a finite set.
• Forgetting the indices in a subsequence must be strictly increasing. $(x_{n_k})$ requires $n_1<n_2<n_3<\cdots$; reusing or reordering indices does not define a valid subsequence.
• Omitting boundedness or closedness when citing Heine-Borel. Both hypotheses are necessary: $(0,1)$ is bounded but not closed (not compact); $\mathbb{Z}$ is closed but not bounded (not compact).
• Believing every bounded sequence converges. Bolzano–Weierstrass only guarantees a convergent subsequence; e.g. $x_n=(-1)^n$ is bounded, does not converge, but has convergent subsequences $x_{2k}\to1$ and $x_{2k+1}\to-1$.

Summary

• Open cover: a collection of open sets whose union contains $K$; compact means every open cover has a finite subcover.
• Exhibiting one open cover with no finite subcover suffices to prove non-compactness; proving compactness directly requires handling all covers.
• Sequentially compact: every sequence has a subsequence converging within $K$; equivalent to compactness in any metric space.
• Bolzano–Weierstrass: every bounded sequence in $\mathbb{R}$ has a convergent subsequence, proved by repeated bisection and the Nested Interval Property.
• Heine-Borel (in $\mathbb{R}$ / $\mathbb{R}^n$): $K$ is compact $\iff$ $K$ is closed and bounded — specific to $\mathbb{R}^n$, not a general metric-space fact.
• A compact subset of any metric space is closed; a closed subset of a compact set is itself compact.