I wish to obtain a quick proof that locally compact Hausdorff spaces are regular. But I am stuck on having to prove that every locally compact Hausdorff space is embedded in a compact Hausdorff space

The definition of locally compact I am using is:

A space $(X, \mathfrak{T})$ is locally compact if $\forall x \in X$, $\exists K, U \subseteq X, K$ is compact, $U$ is open, s.t. $x \in U \subseteq K$

I am completely stuck on this one.

Let $(X,\mathfrak{T})$ be a locally compact Hausdorff space, let $(Y, \mathfrak{J})$ be a compact Hausdorff space, then we wish to exhibit a homeomorphism $f$ such that $f(X) \cong A \subseteq Y$

How can this be done? It seems there is no explicit relationship between the two spaces for me to work with. Help!!

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    $\begingroup$ One way to do this is using the one point compactification (or Alexandrov compatification) of a LCH space, which should be explained in general topology books (Munkres, Willard, etc). $\endgroup$ – Pedro Tamaroff Aug 11 '16 at 5:17
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    $\begingroup$ The one-point compactification is also described here. You may also find this blog post helpful. $\endgroup$ – Brian M. Scott Aug 11 '16 at 6:22
  • $\begingroup$ Do you really need the embedding in a compact space? Since regularity is a local property and $K$ is compact, you are done. $\endgroup$ – egreg Aug 11 '16 at 9:36
  • $\begingroup$ @egreg I think this is what it is done in Coro 1 math.dartmouth.edu/archive/m54x12/public_html/m54lecture20.pdf but I might be wrong $\endgroup$ – Olórin Aug 11 '16 at 9:46

I assume you know the followings results of general topology: (1) if $X$ is compact and Hausdorff then is normal. (2) every normal space is also regular and every subspace of a regular space is regular (3) X is a locally compact space if and only if there exist uniquely a space $X^{\infty}$ (Alexandrov compactification) such that (a) X is a subspace of $X^{\infty}$, (b) the $X^{\infty}$ and $X$ differ of by a point (c) $X^{\infty}$ is compact and Hausdorff. So now you can deduct your statement directly from the previous result: X is locally compact then you can use the result (3) and the result (1) says that $X^{\infty}$ is normal because of point (c) of result (3), then using the (2) argumentation you can say that $X^{\infty}$ is regular and with point (a) of result (3) you can conclude that X is regular.

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    $\begingroup$ $X$ is closed in $X^{\infty}$ if and only if $X$ is compact. In general, $X$ is an open subset of $X^{\infty}$ (assuming $X$ is Hausdorff). $\endgroup$ – egreg Aug 11 '16 at 9:05
  • $\begingroup$ Right, if (b) holds and if X^{\infty} is Hausdorff then the the point is closed, which means that X should be open. Thanks. So do you think there is no way to adjust my proof? Otherwise I will delete it $\endgroup$ – Salvatore Aug 11 '16 at 9:28
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    $\begingroup$ Since $X^{\infty}$ is compact and $X$ embeds in $X^{\infty}$, it follows that $X$ is completely regular (even more than regular), because complete regularity is a hereditary property. $\endgroup$ – egreg Aug 11 '16 at 9:33
  • $\begingroup$ @egreg actually i realised that the question was about regularity and not normality, so there is no need for X to be closed in X^{infty}, since every subspace of a regular space is regular. And normality implies regularity $\endgroup$ – Salvatore Aug 11 '16 at 11:26

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