compactnessgeneral-topology
I've been stuck on this problem for a while:
Prove that a dense subset of a compact Hausdorff space is open iff it is locally compact.
I've solved $(\Longleftarrow)$ (it's actually been posted before) but for the other side I have no idea. I was trying to find a relatively compact open set for each point (in our dense set). That is because we know that X is locally compact iff for each point there is an open and relatively compact set containing it.
Actually, a stronger result is true: If each point has a compact Hausdorff neighborhood, then $X$ is locally compact.
Let $x\in X$, $U$ a neighborhood of $x$, and $K$ a compact Hausdorff neighborhood of $x$. Since $K$ is regular as a subspace of $X$ and $U\cap K$ is a neighborhood of $x$ in $K,$ there is a neighborhood $C\subseteq U\cap K$ of $x$ in $K$ (and thus also in $X$ since $K$ is a neighborhood of $x$) which is closed in $K$, thus compact. This proves that there are arbitrary small compact neighborhoods around $x$.
Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
Best Answer
Let $O$ be an open subset of your $X$. Let $U$ be an open subset of $O$ and $x \in O$. Then $U$ is also open in $X$ and as $X$ is regular (being compact Hausdorff) there is an open subset $V$ of $X$ so that $x \in V \subseteq \overline{V} \subseteq U$. It follows that $\overline{V}$ is a compact neighbourhood of $x$ inside $U$ (and inside $O$) so $O$ is locally compact QED.
The density of the subspace is irrelevant and for $X$ be only need it to be a locally compact Hausdorff space, not even compact.