# Contets

## Statement

###### Proposition

Let $X$ be a topological space which is

1. locally compact (in the sense that every point has an open neighbourhood whose topological closure is compact),

Then $X$ is sigma-compact.

In particular then $X$ is also paracompact since locally compact and sigma-compact spaces are paracompact.

###### Proof

We need to produce a countable cover of $X$ by compact subspaces.

By second-countability there exists a countable base of open subsets

$\beta = \left\{ B_i \subset X \right\}_{i \in I} \,.$

By local compactness, every point $x \in X$ has an open neighbourhood $V_x$ whose topological closure $Cl(V_x)$ is compact.

By definition of base of a topology, there exists $B_x \in \beta$ such that ${x} \subset B_x \subset V_x$, hence $Cl(B_x) \subset Cl(V_x)$. Since $Cl(V_x)$ is compact by assumption, and since closed subspaces of compact spaces are compact it follows that $B_x$ is compact.

Applying this for each point yields that

$X = \underset{x \in X}{\cup} Cl(B_x) \,.$

But since there is only a countable set of base elements $B$ to begin with, there is a countable subset $J \subset X$ such that

$X = \underset{x \in J}{\cup} Cl(B_x) \,.$

Hence

$\{Cl(B_x) \subset X\}_{x \in J}$

is a countable cover of $X$ by compact subspaces.

Revised on June 7, 2017 09:24:26 by Urs Schreiber (185.46.137.5)