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I am trying to figure out how local compactness is behaving under topological constructions. In particular, I am trying to find an example of a locally compact space whose closure is not locally compact but I have not been successful yet.

My definition of local compactness is that for every point $x \in X$ there exists an open neighborhood $U$ and a compact set $K$ such that $$ x \in U \subseteq K.$$

mathology
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    Does locally compact include Hausdorff? If not, what is the definition? Are there requirements concerning the ambient space in which closure is formed? – Paul Frost Aug 10 '19 at 16:00
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    Have you condsidered the particular point topology on an infinite set? Depending on your definition of locally compact it might be satisfactory. – G. Chiusole Aug 10 '19 at 16:24
  • My definition of local compactness is that every point has a compact neighborhood. I will add it to the question. – mathology Aug 10 '19 at 16:44

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Be $(X,\tau_X)$ a space that is not locally compact. Define the ambient space as $A=X\cup\{p\}$ ($p\notin X$), with the topology $\tau_A = \{\emptyset\}\cup\{U\cup\{p\}|U\in\tau_X\}$. Now consider the space $S=\{p\}$.

Clearly, as finite space, $S$ is locally compact. By construction, the only closed set containing $p$ is the complete space $A$, so that is the closed of $S$ in $A$. But $A$ is not locally compact, as any open cover of some set in $X$ can be mapped to an open cover of the set in $A$ by just adding $p$ to each set, and a finite open subcover of that can then be mapped to a finite open subcover in $X$ by simply removing $p$ from each set again. Thus if $A$ were locally compact, then $X$ would be locally compact as well, but we assumed that it isn't, thus $A$ isn't either.

Thus we have a locally compact space $S$ whose closure in $A$ (which is $A$ itself) is not locally compact.

celtschk
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