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 $K\subset K_n$.

• 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 $K_n$.

• 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.

• Not every locally compact space (like $ℝ$) 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 ${\tilde{K}}_n:=\overline{{\bigcup }_{k=1}^n{U}_k}$. Then ${({\tilde{K}}_n)}_{n\in \mathbb{N}}$ 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	2013-03-22 19:08:18
Last modified on	2013-03-22 19:08:18
Owner	karstenb (16623)
Last modified by	karstenb (16623)
Numerical id	8
Author	karstenb (16623)
Entry type	Definition
Classification	msc 54-00
Related topic	SigmaCompact
Defines	hemicompact space