Compact space

A topological space is compact when every open cover has a finite subcover. Every closed subspace of a compact space is compact.

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.

A subset of a finite-dimensional real vector space is compact exactly when it is closed and bounded.

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.

The continuous image of a compact space is compact: pull an open cover of the image back to the domain, take a finite subcover there, and map it forward.

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 $A\cap K$.