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.