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!!
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.