Compact space

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

A topological space is compact if and only if every ultrafilter on it converges to at least one point, assuming the ultrafilter lemma. No Hausdorff assumption is needed. In a compact space, the closures of the filter members have the finite intersection property, so choose a point in their intersection. Every open neighbourhood of that point belongs to the ultrafilter, since otherwise its closed complement would be a filter member. Conversely, extend the complements of a cover with no finite subcover to an ultrafilter; a limit point lies in a covering open set, contradicting the simultaneous membership of that set and its complement.

A compact exhaustion of an open subset $U$ of a finite-dimensional Euclidean space is an increasing sequence of compact subsets $K_n\subset U$ with $K_n\subset \mathrm{int}{K}_{n+1}$ and ${\bigcup }_n{K}_n=U$. For example, choose smoothed bounds or suitable nested levels of distance from the complement and distance from the origin. Exit times from these sets express the lifetime of a maximal local solution of a stochastic differential equation; every finite-time continuous path lying in $U$ has compact image and is contained in some $K_n$.

A compact Hausdorff space is both compact and Hausdorff. Such a space is normal, and every real-valued continuous function on it is bounded.

On a nonempty compact Hausdorff space, let continuous nonnegative functions decrease pointwise. Their suprema decrease to a finite number $L$. For every $t<L$, the sets $\{f_n\ge t\}$ are nonempty nested closed sets; compactness gives a common point with limit value at least $t$. Thus the supremum of the pointwise limit is $L$. The limit need not be continuous, and this conclusion does not claim uniform convergence as in the Dini theorem.

Let continuous maps $f,g:X\rightrightarrows Y$ between compact Hausdorff spaces have an underlying split coequalizer $q:Y\to Z$ with set maps $s,t$ satisfying $qs=1$, $ft=1$ and $gt=sq$. Its equivalence relation is the image of the closed subset $\{(x,x^{\prime })\in X^2:g(x)=g(x^{\prime })\}$ under $(f,f)$. One inclusion follows from $qf=qg$; for the other use $x=t(y)$ and $x^{\prime }=t(y^{\prime })$ whenever $q(y)=q(y^{\prime })$. This image is compact and hence closed in the Hausdorff product $Y^2$. The quotient is therefore compact Hausdorff, giving the preserved coequalizer required by the Beck monadicity theorem.

Embed $\mathbb{N}$ into the compact product $\{0,1{\}}^{\mathcal{P}(\mathbb{N})}$ by $j(n{)}_S=1_S(n)$ and take its closure $X$. The Tychonoff theorem makes $X$ a compact Hausdorff space, while $j(\mathbb{N})$ is countable, dense and discrete in its subspace topology. Yet $X$ is not metrizable. Any convergent subsequence $j(n_k)$ of distinct points would have a convergent coordinate for $S=\{n_2,n_4,\dots \}$, whereas that coordinate alternates between zero and one. This contradicts sequential compactness of a compact metric space. Thus a dense metrizable subspace does not force a separable compact Hausdorff ambient space to be metrizable.

For every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

A family of sets has the finite intersection property when every finite subfamily has nonempty intersection. A topological space is compact exactly when every family of closed subsets with the finite intersection property has nonempty total intersection.

Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

Every quotient of a compact space is compact because the quotient projection is continuous and surjective.

Every open cover of a compact metric space has a number $\delta >0$ such that every subset of diameter less than $\delta$ lies in one member of the cover.

Every compact Hausdorff space is normal: disjoint closed subsets have disjoint open neighbourhoods. Pointwise Hausdorff separation becomes uniform first over one compact closed set and then over the other by taking finite subcovers.

A topological space is compact when every cover by members of a fixed subbase has a finite subcover. Equivalently, every family of complements of subbasic opens with the finite-intersection property has nonempty intersection.

A space is locally compact when every point has a base of neighbourhoods containing compact neighbourhoods. A compact Hausdorff space is locally compact because regular separation produces $x\in V\subseteq \overline{V}\subseteq U$.

If $X$ is locally compact Hausdorff and $A\cap K$ is closed in every compact subset $K$, then $A$ is closed. Around each point outside $A$, use a compact neighbourhood and the relative openness of its complement of $A\cap K$.