# 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 $K\subset K_{n}$.

Let $(X,\tau)$ be a first countable hemicompact space. Then $X$ is locally compact.

###### Proof.

Let $\cdots\subset K_{n}\subset K_{n+1}\subset\cdots$ be an admissible sequence of $X$. Assume for contradiction   that there is an $x\in X$ without compact neighborhood   . Let $U_{n}\supset U_{n+1}\supset\cdots$ 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 HemicompactSpace 2013-03-22 19:08:18 2013-03-22 19:08:18 karstenb (16623) karstenb (16623) 8 karstenb (16623) Definition msc 54-00 SigmaCompact hemicompact space