locally compact and second-countable spaces are sigma-compact



topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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Let XX be a topological space which is

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

  2. second-contable,

Then XX is sigma-compact.

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


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

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

β={B iX} iI. \beta = \left\{ B_i \subset X \right\}_{i \in I} \,.

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

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

Applying this for each point yields that

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

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

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


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

is a countable cover of XX by compact subspaces.

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