Compactness

Let $X$ be a topological space and $A\subseteq X.$ An open cover of $A$ is a collection of open sets whose union contains $A,$ and a subcover is a subcollection of an open cover that still covers $A.$ The set $A$ is compact if and only if every open cover of $A$ has a finite subcover, and $X$ is a compact space if $X$ is compact as a subset of itself. Compactness of $A$ is a property of the subspace $A$ alone, so it does not depend on the space in which $A$ sits.

Finite sets are compact, $\mathbb{R}$ is not compact, since the open cover $\left\{\left(-n,n\right):n\in\mathbb{N}\right\}$ has no finite subcover, and $\left(0,1\right]$ is not compact, since $\left\{\left(\frac{1}{n},2\right):n\in\mathbb{N}\right\}$ has no finite subcover. A closed bounded interval $\left[a,b\right]$ is compact. Compact sets have the following properties:

• a closed subset of a compact set is compact;
• a compact subset of a Hausdorff space is closed;
• a finite union of compact sets is compact;
• the product of finitely many compact spaces is compact;
• the continuous image of a compact set is compact.

The last property states that if $f:X\rightarrow Y$ is continuous and $A\subseteq X$ is compact then $f(A)$ is compact. Consequently a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism, since it maps closed sets, which are compact, to compact sets, which are closed.

A set $A$ in a metric space is sequentially compact if and only if every sequence in $A$ has a subsequence that converges to a point of $A,$ and totally bounded if and only if for every $\varepsilon>0$ it is covered by finitely many open balls of radius $\varepsilon.$ A totally bounded set is bounded, and in $\mathbb{R}^n$ the converse holds. In a metric space the following are equivalent:

• $A$ is compact;
• $A$ is sequentially compact;
• $A$ is complete and totally bounded.

The Heine–Borel theorem states that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. The Bolzano–Weierstrass theorem states that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence, which is the sequential compactness of closed balls. Neither holds in every metric space: the closed unit ball of an infinite-dimensional normed vector space is closed and bounded but not compact, and the discrete metric makes every infinite set closed, bounded and non-compact. The Lebesgue number lemma states that for every open cover of a compact metric space $A$ there exists $\delta>0$ such that every subset of $A$ with diameter less than $\delta$ lies inside some set of the cover.

Let $A$ be a non-empty compact space and $f:A\rightarrow\mathbb{R}$ be continuous. Since $f(A)$ is a compact subset of $\mathbb{R}$ it is closed and bounded, so it contains its supremum and infimum: $f$ is bounded and attains its maximum and minimum.

$$\exists\,x_{\min},x_{\max}\in A:f\left(x_{\min}\right)\leq f(x)\leq f\left(x_{\max}\right)\,\,\forall\,x\in A$$

This extends the extreme value theorem from closed bounded intervals to compact sets, and the Heine–Cantor theorem extends in the same way: a continuous function from a compact metric space to a metric space is uniformly continuous. If $A$ is a compact subset of a metric space and $x\in X$ then the distance $d(x,A):=\inf_{a\in A}d(x,a)$ is attained at some point of $A,$ and two disjoint compact sets are at positive distance from one another.