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As we know a topological space $X$ is said to be locally compact at a point $ x \in X $ if $x$ has a compact neighbourhood in $X$. $X$ is called locally compact if it is locally compact at every point.

Is locally compact space open in any compactification?

Stefan Hamcke
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fatemeh
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1 Answers1

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I take the definition of locally compact saying that each space has a neighborhood base of compact sets. If $X$ is Hausdorff, then this is equivalent to each point having a compact neighborhood.

A compactification is usually required to be a compact Hausdorff space where the original space is densely embedded. Now Hausdorff spaces have the property that a locally compact subspace $S$ can be expressed as the intersection of a closed and an open subset, or equivalently, $S$ is open in $\overline S$. If $X$ is locally compact, then so is its homeomorphic image $i(X)$ in the compactification $\hat X$. As mentioned, $i(X)$ is open in its own closure, which is just $\hat X$, so $i(X)$ is open.

Stefan Hamcke
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  • what do you mean "If X is Hausdorff, then this is equivalent to each point having a compact neighborhood" and "locally compact subspace S can be expressed as the intersection of a closed and an open subset" and "If X is locally compact, then so is its homeomorphic image i(X) in the compactification ?Can you give me more explain?Thanks alot. – fatemeh Aug 12 '13 at 20:17
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    @fatemeh: If you would answer Brian's question, then there may be nothing at all for Stefan to explain. – Cameron Buie Aug 12 '13 at 21:01
  • @fatemeh: One can prove that if a point $x\in X$ has a neighborhood $C$ which is compact and Hausdorff, then $x$ has a neighborhood base of compact sets. So if each point has a compact Hausdorff neighborhood, then the space is locally compact in the sense that each point has a local base of compact sets. – Stefan Hamcke Aug 12 '13 at 21:09
  • For the second point: If $S$ is a locally compact subspace of $X$ and $X$ is Hausdorff, then there is an open set $U$ and a closed set $C$ such that $S=U\cap C.$ I don't really know what I have to explain for the third point. $i$ is the embedding $X\hookrightarrow\hat X.$ – Stefan Hamcke Aug 12 '13 at 21:13