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 $A.$ A subcover of an open cover of $A$ is a subcollection of the open cover whose union still contains $A.$

Let $X$ be a topological space and $A\subseteq X.$ The set $A$ is compact if and only if every open cover of $A$ has a finite subcover. The space $X$ is a compact space if and only 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.

Compact sets have the following properties:

• 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;
• $\left(0,1\right]$ 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;
• 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: if $f:X\rightarrow Y$ is continuous and $A\subseteq X$ is compact then $f(A)$ is compact.

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.

Let $(X,d)$ be a metric space and $A\subseteq X.$ The set $A$ is sequentially compact if and only if every sequence in $A$ has a subsequence that converges to a point of $A.$

Let $(X,d)$ be a metric space and $A\subseteq X.$ The set $A$ is 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.

Let $(X,d)$ be a metric space and $A\subseteq X.$ The following are equivalent:

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

Let $A\subseteq\mathbb{R}^n.$ The Heine–Borel theorem states that $A$ is compact if and only if it is closed and bounded.

Let $\left(\underline{x}_n\right)$ be a bounded sequence in $\mathbb{R}^n.$ The Bolzano–Weierstrass theorem states that $\left(\underline{x}_n\right)$ has a convergent subsequence. The Bolzano–Weierstrass theorem is the sequential compactness of closed balls in $\mathbb{R}^n.$

Neither the Heine–Borel theorem nor the Bolzano–Weierstrass theorem 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.

Let $A$ be a compact metric space with an open cover. The Lebesgue number lemma states that there exists $\delta>0$ such that every subset of $A$ with diameter less than $\delta$ lies inside some set of the cover. Such a $\delta$ is a Lebesgue number of the cover.

Let $A$ be a non-empty compact space and $f:A\rightarrow\mathbb{R}$ be continuous. The image $f(A)$ is a compact subset of $\mathbb{R},$ so it is closed and bounded and contains its supremum and infimum. Hence $f$ 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. The Heine–Cantor theorem extends in the same way: a continuous function from a compact metric space to a metric space is uniformly continuous.

Let $(X,d)$ be a metric space, $A\subseteq X$ be non-empty and $x\in X.$ The distance from $x$ to $A$ is the infimum of the distances from $x$ to points of $A.$

$$d(x,A):=\inf_{a\in A}d(x,a)$$

If $A$ is compact then the distance from $x$ to $A$ is attained at some point of $A,$ and two disjoint compact sets are at positive distance from one another.