# Exhaustion by compact sets

In mathematics, especially general topology and analysis, an exhaustion by compact sets of a topological space $X$ is a nested sequence of compact subsets $K_{i}$ of $X$ (i.e. $K_{1}\subseteq K_{2}\subseteq K_{3}\subseteq \cdots$), such that each $K_{i}$ is contained in the interior of $K_{i+1}$, i.e. $K_{i}\subset {\text{int}}(K_{i+1})$, and $X=\bigcup _{i=1}^{\infty }K_{i}$.
A space admitting an exhaustion by compact sets is called exhaustible by compact sets.
As an example, for the space $X=\mathbb {R} ^{n}$, the sequence of closed balls $K_{i}=\{x:|x|\leq i\}$ forms an exhaustion of the space by compact sets.
There is a weaker condition that drops the requirement that $K_{i}$ is in the interior of $K_{i+1}$, meaning the space is σ-compact (i.e., a countable union of compact subsets.)

## Construction

If there is an exhaustion by compact sets, the space is necessarily locally compact (if Hausdorff). The converse is also often true. For example, for a locally compact Hausdorff space $X$ that is a countable union of compact subsets, we can construct an exhaustion as follows. We write $X=\bigcup _{1}^{\infty }K_{n}$ as a union of compact sets $K_{n}$. Then inductively choose open sets $V_{n}\supset {\overline {V_{n-1}}}\cup K_{n}$ with compact closures, where $V_{0}=\emptyset$. Then ${\overline {V_{n}}}$ is a required exhaustion.
For a locally compact Hausdorff space that is second-countable, a similar argument can be used to construct an exhaustion.

## Application

For a Hausdorff space $X$, an exhaustion by compact sets can be used to show the space is paracompact. Indeed, suppose we have an increasing sequence $V_{1}\subset V_{2}\subset \cdots$ of open subsets such that $X=\bigcup V_{n}$ and each ${\overline {V_{n}}}$ is compact and is contained in $V_{n+1}$. Let ${\mathcal {U}}$ be an open cover of $X$. We also let $V_{n}=\emptyset ,\,n\leq 0$. Then, for each $n\geq 1$, $\{(V_{n+1}-{\overline {V_{n-2}}})\cap U\mid U\in {\mathcal {U}}\}$ is an open cover of the compact set ${\overline {V_{n}}}-V_{n-1}$ and thus admits a finite subcover ${\mathcal {V}}_{n}$. Then ${\mathcal {V}}:=\bigcup _{n=1}^{\infty }{\mathcal {V}}_{n}$ is a locally finite refinement of ${\mathcal {U}}.$
Remark: The proof in fact shows that each open cover admits a countable refinement consisting of open sets with compact closures and each of whose members intersects only finitely many others.
The following type of converse also holds. A paracompact locally compact Hausdorff space with countably many connected components is a countable union of compact sets and thus admits an exhaustion by compact subsets.

## Relation to other properties

The following are equivalent for a topological space $X$:
* $X$ is exhaustible by compact sets.
* $X$ is σ-compact and weakly locally compact.
* $X$ is Lindelöf and weakly locally compact.
(where weakly locally compact means locally compact in the weak sense that each point has a compact neighborhood).
The hemicompact property is intermediate between exhaustible by compact sets and σ-compact. Every space exhaustible by compact sets is hemicompact and every hemicompact space is σ-compact, but the reverse implications do not hold. For example, the Arens-Fort space and the Appert space are hemicompact, but not exhaustible by compact sets (because not weakly locally compact), and the set $\mathbb {Q}$ of rational numbers with the usual topology is σ-compact, but not hemicompact.
Every regular Hausdorff space that is a countable union of compact sets is paracompact.