If regular includes Hausdorff, the answer is given [here](http://math.stackexchange.com/q/49727) (without using regularity), else you should probably clarify what "locally compact" means, because there are various non-equivalent formulations. – 2012-04-18
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Would $\mathbb{Q}$ as a subset of $\mathbb{R}$ be a counterexample? Perhaps I misunderstand what a locally compact subspace is... – 2012-04-18
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@copper.hat: a locally compact subspace is a subspace which is locally compact in the induced topology. $\mathbb{Q}$ with the subspace topology of $\mathbb{R}$ isn't locally compact e.g. by Baire (it is a countable union of nowhere dense sets - points). The statement is true for locally compact subspaces of a Hausdorff space, as shown in the thread I linked to in my previous comment. – 2012-04-18
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@copper.hat $\mathbb{Q}$ is not locally compact. For it to be locally compact at point $x$ there must be a compact subset $S$ of $\mathbb{Q}$ containing an open neighborhood of $x$. But then $S$ must contain $A=[a,b]\cap\mathbb{Q}$ for some $a,b$, and $A$ is closed in $S$, therefore, compact as well. But $A$ is not compact: for a metric space compactness and sequential compactness are the same, now take a sequence of points in $A$ converging to an irrational number. – 2012-04-18
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Thanks both of you! ! – 2012-04-18
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@t.b. if X is hausdorff (regular) the following statement are equivalent: 1. X each point in X hast at least one compact nhood 2.each point In X has a nhoob base consisting of compact nhoods 3.X is a strongly locally compact. and since every subspace of aregular space is regular ,These three propositions can also be used instead. – 2012-04-18
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[49727] Dense and locally compact subset of a Hausdorff space is open