Exhaustion by compact sets

In mathematics, especially general topology and analysis, an exhaustion by compact setsTemplate:Sfn 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 $X=\bigcup _{i=1}^{\mathrm{\infty }}{K}_i$.

A space admitting an exhaustion by compact sets is called exhaustible by compact sets.Template:Sfn

As an example, for the space $X={ℝ}^n$, the sequence of closed balls $K_i=\{x:|x|\le i\}$ 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 $X=\bigcup _1^{\mathrm{\infty }}{K}_n$ 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 $V_0=\mathrm{\varnothing }$. 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 $𝒰$ be an open cover of $X$. We also let $V_n=\mathrm{\varnothing },n\le 0$. Then, for each $n\ge 1$, $\{(V_{n+1}-\overline{V_{n-2}})\cap U\mid U\in 𝒰\}$ is an open cover of the compact set $\overline{V_n}-V_{n-1}$ and thus admits a finite subcover ${𝒱}_n$. Then $𝒱:=\bigcup _{n=1}^{\mathrm{\infty }}{𝒱}_n$ is a locally finite refinement of $𝒰.$

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 $X$:

1. $X$ is exhaustible by compact sets.
2. $X$ is σ-compact and weakly locally compact.
3. $X$ 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.Template:Fact

References

• Leon Ehrenpreis, Theory of Distributions for Locally Compact Spaces, American Mathematical Society, 1982. Template:ISBN.
• Hans Grauert and Reinhold Remmert, Theory of Stein Spaces, Springer Verlag (Classics in Mathematics), 2004. Template:ISBN.
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External links

• Template:Planetmath reference
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