I have seen two definitions of locally compact topological space

  • A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists a compact set that contains an open neighborhood of $x$.
  • A topological space $(X,\mathcal{F})$ is said to be locally compact if for every $x\in X$, there exists an open neighborhood of $x$ whose closure is compact.

Are these two definitions equivalent? It's clear that the second implies the first, but I don't see why the first implies the second. Suppose that $U $ is an open neighborhood of $x$ with $x\in U\subset K$, here $K$ is a compact set. Since compact sets may not be closed in a non-Hasudorff topological space, the closure of $U$ may not be contained $K$, so we can't say the closure of $U$ is compact.

  • $\begingroup$ just take the closure of the open neighborhood from the first. A closed subset of a compact set is always compact. $\endgroup$ – Forever Mozart Dec 14 '15 at 3:25
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    $\begingroup$ @ForeverMozart If $U\subset C$, and $C$ is compact, then without the Hausdorff condition that is no reason that we will have $\overline{U} \subset C$. $\endgroup$ – Slade Dec 14 '15 at 3:28

Let $X$ be a line with infinitely many origins, i.e. the space obtained by gluing together infinitely many copies of $\mathbb{R}$ along the open subset $\mathbb{R}\setminus\{0\}$.

Each point of $X$ has an open neighborhood homeomorphic to $\mathbb{R}$, so $X$ satisfies definition 1. But the closure of any neighborhood of one of the origins cannot be compact, since it contains an infinite discrete set as a closed subspace, so $X$ does not satisfy definition 2.

We could also glue together infinitely many copies of the Sierpinski space for a similar counterexample.

  • $\begingroup$ Thanks for your answer. I meet the same question in the definition of relatively compact set. (i) A set is relatively compact if it is contained in a compact set. (ii) A set is relatively if its closure is compact. These two definitions are also not equivalent? $\endgroup$ – Xiang Yu Dec 14 '15 at 3:37
  • $\begingroup$ @XiangYu Correct. The same example works, and for the same reason: any one of the origins has a compact neighborhood (e.g. the corresponding $[-1,1]$), but the closure of any such neighborhood must include all the origins, hence not be compact. So this neighborhood is relatively compact in the first sense, but not the second. $\endgroup$ – Slade Dec 14 '15 at 4:10

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