# 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.

• What does core-compact mean? – Stefan Hamcke Jan 27 '14 at 23:22
• Every neighborhood $U$ of every point $x$ contains a neighborhood $V$ of $x$ such that every open cover of $U$ contains a finite subcover of $V$. – Robert Wolfe Jan 27 '14 at 23:23
• Hmm... Wouldn't the closure of this $V$ be compact? Since if $(W_i)_I$ is an open cover of $\bar V$, then $(W_i)_I\cup\{U-\bar V\}$ is an open cover of $U$. Then it contains a finite subcover of $V$, but $U-\bar V$ is not part of this subcover, so the subvocer must also cover $\bar V$. – Stefan Hamcke Jan 27 '14 at 23:29
• In the comment above, I agree that the set $U-\overline{V}$ is not part of the subcover which covers $V$, but how do you conclude then that $\overline{V}$ is still contained in that same subcover though? – Robert Wolfe Jan 28 '14 at 2:25
• If we start with everything in the original post, regularity would imply there is a closed set $x\in C\subseteq V$ with nonempty interior. Take an open cover of $C$ and add $C^c$. That's an open cover of $U$ containing a finite subcover of $V$. And we can remove $C^c$. – Robert Wolfe Jan 28 '14 at 3:18

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 $C$.

• Why couldn't $x\in\overline{U''}$? In addition to that, I'm confused why you introduced sequences. We do not know that our space is even sequential. – Robert Wolfe Feb 5 '14 at 8:38
• Actually, you only need local regularity so it simplifies. – SomeEE Feb 5 '14 at 16:20
• One final thing. I follow how you got that points and closed sets (closed in $X$) in $V$ can be separated by open neighborhoods, but how does that get you a closed set (closed in $X$) $C\subseteq V$ with $z\in C$? I know that regularity can be formulated in two different ways (as neighborhoods or closed sets), but does that extend to local regularity? – Robert Wolfe Feb 5 '14 at 17:28
• "Every neighborhood U of z contains a V ...." is the definition of core-compact. Therefore we may assume that U is regular and C is closed in U which is all that we need. – SomeEE Feb 5 '14 at 18:31

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.

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 $U \subseteq W \subseteq V$.