Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 S 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. 2