Proof help. Core-compactness, Hausdorff, Locally Compact While reading about topologies on continuous function spaces, I've seen remarks that core-compact and locally compact are equivalent for Hausdorff spaces.
Now I can clearly see that locally compact always implies core-compact, so the Hausdorff condition comes into the proof of the converse.
Let $K$ be a core-compact Hausdorff space. Let $x\in U\subseteq K$ with $U$ open. I need to show that $U$ contains a compact neighborhood of $x$. Since $K$ is core-compact there is an open neighborhood $V$ such that $x\in V\subseteq U$ with $V\ll U$.
I don't know where to go from here.
EDIT
Here are the two papers which make make me believe this is a theorem:
Core Compactness and Diagonality in Spaces of Open Sets
Topologies on Spaces of Continuous Functions
I have found another source which claims something better: every sober core-compact space
is locally compact (although I can't see the proof of the theorem; I might buy it).
Non-Hausdorff Topology
This hints that the property of Hausdorff spaces that we want to exploit is the fact that the intersection of all closed neighborhoods of a point is precisely that point.
EDIT
I am now asking for help in completing the proof. This has been bothering me for too long.
 A: This can be done easily with ultrafilters. Similarly to the ultrafilter characterisation of compactness, $V\ll U$ is equivalent to saying that each ultrafilter containing $V$ has a limit in $U$. 
For the proof, let $x\in X$. By core-compactness, we can choose open neighborhoods $U$ and $V$ of $x$ such that $V\ll U$ and $U\ll X$. All ultrafilters containing $V$ have no limit outside of $U$, since they already have a limit in $U$, and they can only have one limit thanks to Hausdorffness. But this means that $\overline{V}\subseteq U$. Hence every ultrafilter containing $\overline{V}$ contains $U$, and since $U\ll X$, these ultrafilters converge. As $\overline{V}$ is closed, they converge to something in $\overline{V}$. So $\overline{V}$ is compact. So $x$ has a compact neighborhood.
A: This is probably not the quickest but here is a suggestion for core-compact $\Rightarrow$ local regularity.
Let $z \in X$.  Let $z \in V \subset U$ as in the definition of core compact.  Let $x \in V$ and $E \subset V$ a closed subset not containing $x$.  For each $y \in F=\overline{E}^U$ choose an open neighborhood $U_y$ with $x \notin \overline{U_y}$ which we may do because $X$ is Hausdorff. The cover $\{U_{y}\}_{y \in F} \cup \{ F^c =U \setminus F \}$ is an open cover of $U$ and so has a finite subcover $U_{y_1},...,U_{y_n}, F^c$ which covers $V$.
Set $U' = U_{y_1} \cup U_{y_2} \cup ... \cup U_{y_n}$. The open set $U'$ covers $E$.  
$x \notin \overline{U_{y_1}},...,\overline{U_{y_n}}$ by construction of the open sets $U_y$ so $x \notin \overline{U'}$.  That is to say, $V$ is regular.
Rest of proof from the comment section added for completeness:
First use local regularity to reduce to the case when $U$ is regular.
Let $z \in X$ and $z \in V \subset U$ as in the definition of core compact.  $U$ is regular by assumption so there exists a closed neighborhood $C \subset V$ containing $z$.  Any open cover of $C$ extends to an open cover of $U$ by adding in $U \setminus C$.  Any such open cover has a finite subcover covering $V$ which covers $C$.  Removing the set $U \setminus C$ from this finite subcover yields a finite subcover of $C$.
A: This would be better suited as a comment but I cannot yet do that.
You would probably find a proof online quite easily if you searched for "exponentiability" but for a more specific proof go for "A Compendium of Continous Lattices" ( You can find this online ) by Gierz, Hofmann, Keimel, Lawson, Mislove, Scott. I believe there is shown straightforwardly something like given Hausdorff space and $U, V$ open then $U \ll V$ implies existence of compact $W$ so that $U \subseteq W \subseteq V$.
