Locally compact and hausdorff proof

[ForMyThunder]

Homework Statement

Let $$X$$ be a locally compact, Hausdorff topological space. If $$x$$ is an element of $$X$$ and $$U$$ is a neighborhood of $$x$$, find a compact neighborhood of $$x$$ contained in $$U$$.

Homework Equations

The Attempt at a Solution

Let $$N$$ be a compact neighborhood of $$x_$$. The set $$D=Fr(N\cap\bar U)$$ is closed, hence compact. For each $$y\in D$$, there exist disjoint neighborhoods $$N_y$$ and $$N_y'$$ of $$y$$ and $$x$$, respectively. The set $$\{N_y:y\in D\}$$ is an open cover of $$D$$, hence it has a finite subcover $$\{N_{y_n}:y_n\in D\}$$. The set $$\cap \bar N_{y_n}'}\subset N\cap U$$ is a closed neighborhood of $$x$$, hence it is compact.

Is this correct?

 

[micromass]

How can you be certain that $$\bigcap \overline{N_{y_n}^\prime}$$ is a subset of $$N\cap U$$?? I think you still need to take the intersection with $$\overline{U}^\prime$$...

 

[ForMyThunder]

I think so. The N'_y's are contained in U^0, and that every neighborhood of D is disjoint from the intersection of the N'_y's shows that their closure is also disjoint from it, and contained in U^0. Closed neighborhood contained in N => compact neighborhood.

This was my thinking, and its the part I'm worried about. If you've got any hints to an easier way...

^0 denotes interior.