Given a locally compact Hausdorff space.

Every compact set has a compact neighborhood base: $$C\subseteq U:\quad N\subseteq U\quad(C\subseteq N^°)$$

My construction was contrary to Rudin's: $$N_c\subseteq U:\quad C\subseteq\bigcup_{c\in C}N_c^°\implies C\subseteq N_1^°\cup\ldots N_n^°$$

But I needn't make use of the Hausdorff property; or did I miss something?

(Hausdorff seems rather to guarantee that compact sets are closed.)

  • $\begingroup$ It depends on which definition of local compactness he’s using. $\endgroup$ – Brian M. Scott Jan 17 '15 at 20:24
  • $\begingroup$ @BrianM.Scott: Aha, right he uses only the first one. $\endgroup$ – C-Star-W-Star Jan 17 '15 at 20:34

This answer is just to check off the thread.

As Brian M. Scott rightly observed the problem is that Rudin uses a weaker definition.


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