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Is it true that in a locally compact (not necessarily Hausdorff) space, every point has a neighbourhood bases consisting of precompact subsets? (precompact means closure is compact).

In addition, if the space is also Hausdorff, then we have a local base consisting of compact neighbourhoods.

The first assertion is easy to see in a Hausdorff space as compact subsets of a Hausdorff space is closed. However, I cannot see this without the Hausdorff condition.

Locally compact means every point has a neighbourhood (a set containing an open set containing that point) which is compact.

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    $\begingroup$ What’s your definition of local compactness? There are several in use. $\endgroup$ – Brian M. Scott May 26 '13 at 5:07
  • $\begingroup$ And indeed the Wikipedia page linked by @BrianM.Scott answers your question. $\endgroup$ – kahen May 26 '13 at 5:07
  • $\begingroup$ I added the definition of locally compact. $\endgroup$ – Vishal Gupta May 26 '13 at 5:11
  • $\begingroup$ Could you be more specific about the answer. I did not get it on that page. $\endgroup$ – Vishal Gupta May 26 '13 at 5:21
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Let $X$ be an infinite set, $p \in X$ fixed, and as a topology on $X$ we use the included point topology on $X$: $O$ is open iff $O = \emptyset$ or $p \in O$. $X$ is not compact, as the cover by all sets $\{x,p\}, x \neq p$, shows: we cannot spare a single element.

Also, the closed sets are $X$ and all sets missing $p$. This means that the closure of every non-empty open set equals $X$, as every non-empty open set contains $p$ and the only closed set that contains $p$ is $X$. So there are no precompact neighbourhoods at all. But $X$ is locally compact according to your definition, as every point $x$ has a finite (hence compact) open neighbourhood $\{x,p\}$.

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  • $\begingroup$ Exactly, however, I wanted to make sure as this was mentioned in passing in the book "Principles of Harmonic Analysis" by "Anton Deitmar" Page 7. ISBN: 978-0-387-85468-7 $\endgroup$ – Vishal Gupta May 27 '13 at 3:29

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