Exhaustion by compact sets explained

X

is a nested sequence of compact subsets

Ki

of

X

(i.e.

K1\subseteqK2\subseteqK3\subseteq

), such that each

Ki

is contained in the interior of

Ki+1

, i.e.

Ki\subsetint(Ki+1)

, and
infty
X=cup
i=1

Ki

.

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

As an example, for the space

X=Rn

, the sequence of closed balls

Ki=\{x:|x|\lei\}

forms an exhaustion of the space by compact sets.

There is a weaker condition that drops the requirement that

Ki

is in the interior of

Ki+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=

infty
cup
1

Kn

as a union of compact sets

Kn

. Then inductively choose open sets

Vn\supset\overline{Vn-1

} \cup K_n with compact closures, where

V0=\emptyset

. Then

\overline{Vn}

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

V1\subsetV2\subset

of open subsets such that

X=cupVn

and each

\overline{Vn}

is compact and is contained in

Vn+1

. Let

l{U}

be an open cover of

X

. We also let

Vn=\emptyset,n\le0

. Then, for each

n\ge1

,

\{(Vn-\overline{Vn-2

}) \cap U \mid U \in \mathcal \} is an open cover of the compact set

\overline{Vn}-Vn-1

and thus admits a finite subcover

l{V}n

. Then

l{V}:=

infty
cup
n=1

l{V}n

is a locally finite refinement of

l{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[1] and thus admits an exhaustion by compact subsets.

Relation to other properties

The following are equivalent for a topological space

X

:[2]

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[3] 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),[4] and the set

\Q

of rational numbers with the usual topology is σ-compact, but not hemicompact.[5]

Every regular Hausdorff space that is a countable union of compact sets is paracompact.

References

External links

Notes and References

  1. NB: the proof in the reference looks problematic. It can be fixed by constructing an open cover whose member intersects only finitely many others. (Then we use the fact that a locally finite connected graph is countable.)
  2. Web site: A question about local compactness and $\sigma$-compactness . Mathematics Stack Exchange.
  3. Web site: Does locally compact and $\sigma$-compact non-Hausdorff space imply hemicompact? . Mathematics Stack Exchange.
  4. Web site: Can a hemicompact space fail to be weakly locally compact? . Mathematics Stack Exchange.
  5. Web site: A $\sigma$-compact but not hemicompact space? . Mathematics Stack Exchange.