hemicompact space
A topological space^{} $(X,\tau )$ is called a hemicompact space if there is an admissible sequence in $X$, i.e. there is a sequence of compact sets ${({K}_{n})}_{n\in \mathbb{N}}$ in $X$ such that for every $K\subset X$ compact there is an $n\in \mathbb{N}$ with $K\subset {K}_{n}$.

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The above conditions imply that if $X$ is hemicompact with admissible sequence ${({K}_{n})}_{n\in \mathbb{N}}$ then $X={\bigcup}_{n\in \mathbb{N}}{K}_{n}$ because every point of $X$ is compact and lies in one of the ${K}_{n}$.

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A hemicompact space is clearly $\sigma $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 $\mathbb{Q}$ with the induced euclidean topology. $\mathbb{Q}$ is $\sigma $compact but not hemicompact. Since $\mathbb{Q}$ satisfies the first axiom of countability it can’t be hemicompact as this would imply local compactness.

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Not every locally compact space (like $\mathbb{R}$) is hemicompact. Take for example an uncountable discrete space. If we assume in addition $\sigma $compactness we obtain a hemicompact space (see below).
Proposition^{}. Let $(X,\tau )$ be a first countable hemicompact space. Then $X$ is locally compact.
Proof.
Let $\mathrm{\cdots}\subset {K}_{n}\subset {K}_{n+1}\subset \mathrm{\cdots}$ be an admissible sequence of $X$. Assume for contradiction^{} that there is an $x\in X$ without compact neighborhood^{}. Let ${U}_{n}\supset {U}_{n+1}\supset \mathrm{\cdots}$ be a countable basis for the neighbourhoods of $x$. For every $n\in \mathbb{N}$ choose a point ${x}_{n}\in {U}_{n}\setminus {K}_{n}$. The set $K:=\{{x}_{n}:n\in \mathbb{N}\}\cup \{x\}$ is compact but there is no $n\in \mathbb{N}$ with $K\subset {K}_{n}$. We have a contradiction. ∎
Proposition. Let $(X,\tau )$ be a locally compact and $\sigma $compact space. Then $X$ is hemicompact.
Proof.
By local compactness we choose a cover $X\subset {\bigcup}_{i\in I}{U}_{i}$ of open sets with compact closure (take a compact neighborhood of every point). By $\sigma $compactness there is a sequence ${({K}_{n})}_{n\in \mathbb{N}}$ of compacts such that $X={\bigcup}_{n\in \mathbb{N}}{K}_{n}$. To each ${K}_{n}$ there is a finite subfamily of ${({U}_{i})}_{i\in I}$ which covers ${K}_{n}$. Denote the union of this finite family by ${U}_{n}$ for each $n\in \mathbb{N}$. Set ${\stackrel{~}{K}}_{n}:=\overline{{\bigcup}_{k=1}^{n}{U}_{k}}$. Then ${({\stackrel{~}{K}}_{n})}_{n\in \mathbb{N}}$ is a sequence of compacts. Let $K\subset X$ be compact then there is a finite subfamily of ${({U}_{i})}_{i\in I}$ covering $K$. Therefore $K\subset {K}_{n}$ for some $n\in \mathbb{N}$. ∎
Title  hemicompact space 

Canonical name  HemicompactSpace 
Date of creation  20130322 19:08:18 
Last modified on  20130322 19:08:18 
Owner  karstenb (16623) 
Last modified by  karstenb (16623) 
Numerical id  8 
Author  karstenb (16623) 
Entry type  Definition 
Classification  msc 5400 
Related topic  SigmaCompact 
Defines  hemicompact space 