Compactness

The deepest theorems of single-variable analysis hold on closed bounded intervals: a continuous function on $[a,b]$ attains its extremes, is uniformly continuous, and is Riemann integrable. Compactness is the property behind all three, isolated so it can be stated in any metric space. It has three equivalent formulations, a covering condition, a sequential condition, and a finiteness-plus-completeness condition, which coincide in a metric space. The Heine–Borel theorem then identifies the compact subsets of ${\mathbb{R}}^n$ as exactly the closed bounded ones.

The covering definition

The quantifier order is the whole content: no matter how the cover is chosen, however large or pathological, finitely many of its sets already suffice. This turns infinitely many local facts (each point sits in some $U_{\lambda }$) into one global fact (a fixed finite list covers everything), which is why compactness converts pointwise information into uniform information.¹

First examples. Finiteness of the subcover is a strong demand.

• $\{0\}$ and every finite set are compact: pick one cover-set through each of the finitely many points.
• $\mathbb{R}$ is not compact: the cover $\{(-j,j){\}}_{j\in \mathbb{N}}$ has no finite subcover, since a finite union is the largest single interval $(-j_m,j_m)$, missing $j_m$.
• $(0,1)$ is not compact: for the cover $\{(1/j,1-1/j)\}$, a finite union is the largest single interval $(1/j_m,1-1/j_m)$, which misses points near the endpoints.

The two failures are different: $\mathbb{R}$ is unbounded, $(0,1)$ is bounded but not closed. Either property rules out compactness, and in ${\mathbb{R}}^n$ they are the only obstructions.

Compact implies closed and bounded

Bounded. Fix $p\in X$. The balls $\{B(p,n){\}}_{n\in \mathbb{N}}$ cover $K$ (they cover all of $X$), so a finite subcover puts $K$ inside a single $B(p,n_m)$.

Closed. If some $x\in \overline{K}\setminus K$, the complements of the closed balls $C(x,1/n)$ form an open cover of $K$ (their union omits only $x$), but any finite subfamily reduces to a single $C(x,1/n_m{)}^c$ that still misses points of $K$ near $x$, since $x\in \overline{K}$. No finite subcover exists, so $K$ cannot be compact unless it is closed.

The converse, closed and bounded implies compact, is false in general and holds only under extra structure. Two counterexamples show where it breaks:

• In $C[0,1]$ the closed unit ball is closed and bounded but not compact: a sequence of functions with $d(f_n,0)=1$ and $d(f_n,f_k)=1$ for $n\ne k$ has no convergent subsequence.
• Under the discrete metric on an infinite set, every subset is closed and bounded, yet only the finite subsets are compact.

Bounded is too weak in infinite-dimensional or discrete spaces; the right strengthening is total boundedness, below.

Sequential compactness

The extreme value theorem rests on a condition about sequences, not covers, and in a metric space it is equivalent to compactness.

One direction: if $K$ is compact and some sequence $\{x_n\}$ had no convergent subsequence in $K$, then each $x\in K$ would have a ball containing only finitely many terms; these balls cover $K$, a finite subfamily contains only finitely many terms in total, yet must contain all of $K$ and hence all terms, a contradiction. The other direction runs through a covering lemma that produces a single radius working uniformly across a sequentially compact set.

The uniform radius $\delta$ (depending on the cover, not on $x$) lets one pick points $x_1,x_2,\dots$ in $K$ with pairwise distance at least $\delta$, each $B(x_j,\delta )$ inside a cover-set; by sequential compactness the selection cannot continue forever, so the process halts with a finite subcover.

Total boundedness and the three-way equivalence

The correct strengthening of bounded is finiteness at every scale.

Total boundedness requires that $S$ be coverable by finitely many balls of any prescribed radius, a far stronger condition than fitting inside one large ball. It rules out the infinite discrete set (balls of radius $\frac{1}{2}$ are singletons, and infinitely many are needed) and the $C[0,1]$ unit ball. With it, compactness in a metric space has three equivalent descriptions.

Completeness and total boundedness separate the two failure modes cleanly. $(0,1)$ is totally bounded but not complete; an infinite discrete space is complete but not totally bounded; each is missing exactly one half of compactness. In a complete space the theorem specializes: $K$ is compact iff it is closed and totally bounded, because a closed subset of a complete space is itself complete.

The Heine–Borel theorem

In ${\mathbb{R}}^n$, bounded sets are automatically totally bounded (a box is tiled by finitely many small subboxes), so the extra hypothesis is free and the familiar criterion holds.

The forward direction is the general proposition above. For the converse, a closed bounded $K\subseteq {\mathbb{R}}^n$ sits inside a box $[a_1,b_1]\times \cdots \times [a_n,b_n]$; repeatedly extracting convergent subsequences coordinate by coordinate, using Bolzano–Weierstrass in each, produces a subsequence converging in the box, and closedness keeps the limit in $K$. So $K$ is sequentially compact, hence compact.

Heine–Borel is a theorem about ${\mathbb{R}}^n$, resting on Bolzano–Weierstrass and thus on the completeness of $\mathbb{R}$. It does not extend to general metric spaces, nor even to every subspace of ${\mathbb{R}}^n$; the $C[0,1]$ unit ball is the standard counterexample showing that closed and bounded is not enough once the dimension is infinite.

A worked compact set

Nested compact sets and the finite intersection property

Compactness has a dual phrasing in terms of closed sets, obtained by complementing the cover definition.

Complementing a would-be cover with no finite subcover produces closed sets with the finite intersection property but empty intersection, and conversely. The most-used consequence is a Cantor-style nesting result.

The $K_n$ are closed subsets of the compact $K_1$, and any finite subfamily intersects in the smallest listed set, which is nonempty; the finite intersection property then forces the full intersection to be nonempty. This is the compact-set counterpart of the nested closed balls in a complete space, and it underlies existence proofs that produce a point as the sole common element of a shrinking family.

Stability and use

Compactness passes to closed subsets and does not depend on the ambient space.

The last point contrasts sharply with closedness, which depends on the ambient space; compactness does not, so one may speak of a compact space without naming where it sits. This makes compactness the right hypothesis for the strongest continuity theorems: the continuous image of a compact set is compact, which yields the extreme value theorem, and continuity on a compact set is automatically uniform.