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hemicompact space∗
karstenb†
2013-03-22 3:19:46
A topological space (X, τ ) is called a hemicompact space if there is an admissible sequence in X, i.e. there is a sequence of compact sets (Kn )n∈N in X
such that for every K ⊂ X compact there is an n ∈ N with K ⊂ Kn .
• The above conditions implySthat if X is hemicompact with admissible
sequence (Kn )n∈N then X = n∈N Kn because every point of X is compact
and lies in one of the Kn .
• A hemicompact space is clearly σ-compact. The converse is false in general. This follows from the fact that a first countable hemicompact space is
locally compact (see below). Consider the set of rational numbers Q with
the induced euclidean topology. Q is σ-compact but not hemicompact.
Since Q satisfies the first axiom of countability it can’t be hemicompact
as this would imply local compactness.
• Not every locally compact space (like R) is hemicompact. Take for example an uncountable discrete space. If we assume in addition σ-compactness
we obtain a hemicompact space (see below).
Proposition. Let (X, τ ) be a first countable hemicompact space. Then X
is locally compact.
Proof. Let · · · ⊂ Kn ⊂ Kn+1 ⊂ · · · be an admissible sequence of X. Assume
for contradiction that there is an x ∈ X without compact neighborhood. Let
Un ⊃ Un+1 ⊃ · · · be a countable basis for the neighbourhoods of x. For every
n ∈ N choose a point xn ∈ Un \ Kn . The set K := {xn : n ∈ N} ∪ {x} is compact
but there is no n ∈ N with K ⊂ Kn . We have a contradiction.
Proposition. Let (X, τ ) be a locally compact and σ-compact space. Then
X is hemicompact.
∗ hHemicompactSpacei created: h2013-03-2i by: hkarstenbi version: h42037i Privacy
setting: h1i hDefinitioni h54-00i
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Proof. By local compactness we choose a cover X ⊂ i∈I Ui of open sets
with compact closure (take a compact neighborhood of every point).
S By σcompactness there is a sequence (Kn )n∈N of compacts such that X = n∈N Kn .
To each Kn there is a finite subfamily of (Ui )i∈I which covers Kn . Denote the
Sn
union of this finite family by Un for each n ∈ N. Set K̃n := k=1 Uk . Then
(K̃n )n∈N is a sequence of compacts. Let K ⊂ X be compact then there is a
finite subfamily of (Ui )i∈I covering K. Therefore K ⊂ Kn for some n ∈ N.
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