0
$\begingroup$

A topological space is locally compact then there is an open base at each point has all of its set with compact closure, how can i prove it?

It seems very meaningful when the topological space is metrizable and the topology is the topology induced by the metric, but i could not prove it for general topological spaces.

I am not familiarized with Hausdorff spaces.

What i tried for now is starting from the definition of locally compact as the topological space whose ever point has a neighbourhood with compact closure. I picked (for each point) this neighbourhood, and created a class with all open sets contained in that neighbourhood, then i thought that this class may be the open base at the analized point whose all sets have compact closure, but i could not prove that this class is an open base at point, have someone an idea? Or other path to prove it, or yet is this claiming false?

  • 0
    In a non-Hausdorff space there can be a difference between "having a compact neighbourhood" and "having an open neighbourhood with compact closure"; the latter implying the former. The latter automatically transfers having one such neighbourhood to having a local base of them,as we see here. – 2017-01-17

1 Answers 1

1

The idea is right: suppose $x \in X$ has a neighbourhood $U_x$ with the property that $\overline{U_x}$ is compact. Define $\mathcal{V} = \{ V \subseteq U_x: V \text{ open and } x \in V \}$,as you did. Then for $V \in \mathcal{V}$ we have that $\overline{V} \subseteq \overline{U_x}$, and as a closed subset of a compact set is compact (in any space), the members of $\mathcal{V}$ have compact closures.

That it is a base at $x$ is also clear: take any open $O$ with $x \in O$. Then $V = U_x \cap O \in \mathcal{V}$, as it is open, contains $x$ and is a subset of $U_x$ and $x \in V \subseteq O$ trivially as well. So $\mathcal{V}$ is a local base at $x$, as required.