# Locally compact Hausdorff space

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• Apr 8th 2011, 10:44 PM
bubble86
Locally compact Hausdorff space
In a locally compact hausdorff space X for every x in X there exists a compact set that contains an open set V s.t. x is in V. I have to prove this is equivalent to for every open set U s.t. x is in U, there exists a open set V s.t x is in V and closure of V is compact and contained in U. I know how to prove it using the one point compactification of the space X, is there an easier way to prove this. for the reverse implication is easy. any hints would be helpful
• Apr 9th 2011, 09:07 AM
hatsoff
Use the theorem that compact subspaces of a Hausdorff space are always closed. In particular, if $A$ is compact and $O\subseteq A$ is open, then $\overline{O}\subseteq\overline{A}=A$. You need an open neighborhood whose closure is contained in another open neighborhood. So just find an open neighborhood which includes the compact neighborhood guaranteed by local compactness, and another open neighborhood included in that compact neighborhood.
• Apr 9th 2011, 11:10 AM
bubble86
say U is an open n-hood about x and O is my open n-hood about x that is contained in K a compact set. pick a basis element, B of the topolgy that contains x and B is contained in the intersection of O and U. now how can i guarantee that closure of B, which is compact, is contained in U.
• Apr 9th 2011, 11:15 AM
hatsoff
Yup, that will work.

Although remember that finite intersections of open sets are open, so there's no need to drag bases into the argument if you don't want to.
• Apr 9th 2011, 11:39 AM
bubble86
hey sorry but how would i ensure that the closure of B is contained in the given open n-hood, U.
• Apr 9th 2011, 12:21 PM
hatsoff
Sorry, I misread your OP. This is not as simple as it seemed. My apologies.

I'll try to get a REAL solution later on.

EDIT:

Okay, here we are...

Let $x\in X$ and $U$ an open neighborhood of $x$. By the local compactness condition, $x$ has a compact neighborhood $K$ with an open neighborhood $O\subseteq K$. Notice that $K\cap U^c$ is closed by the Hausdorff condition and therefore compact (since closed subspaces of a compact space are compact). Since $x\notin K\cap U^c$, there is a theorem which tells us that $K\cap U^c$ compact gives us disjoint open sets $P,Q$ by the Hausdorff condition with $x\in P$ and $K\cap U^c\subseteq Q$ (this is easy to show). So $Q^c\subseteq (K\cap U^c)^c$ and therefore $$Q^c\cap K\subseteq (K\cap U^c)^c\cap K=U\cap K\subseteq U$$. So $V:=P\cap O\subseteq Q^c\cap K$ is an open neighborhood with $\overline{V}\subseteq\overline{Q^c\cap K}=Q^c\cap K$ (again by the Hausdorff condition).
• Apr 9th 2011, 12:24 PM
bubble86
not quite what i am looking for. what i want is closure of V is contained in U.
• Apr 10th 2011, 12:05 AM
Drexel28
Quote:

Originally Posted by bubble86
not quite what i am looking for. what i want is closure of V is contained in U.

hatsoff gave the correct proof, you must have misread it.
• Apr 10th 2011, 05:47 AM
hatsoff
Quote:

Originally Posted by Drexel28
hatsoff gave the correct proof, you must have misread it.

I gave an incorrect proof, then noticed that I misread his OP, then fixed it.

Oh well.