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How can I prove the following theorem?

Let $X$ locally compact, $K \subseteq X$ compact, $O \subseteq X$ open with $K \subseteq O$. Then there exists $f \in C_c(X)$ with $0 \leq f \leq 1$, $f|_K \equiv 0$ and $\operatorname{supp}(f) \subseteq O$.

(Where $C_c(X)=\{f ∈ C(X, \mathbb C) : \operatorname{supp}(f) \text{ compact}\}$).

I think that I would need the topological lemma for locally compact spaces, but I couldn't come up with a suitable proof. Thanks in advance!

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You can see the proof of Urysohn's Lemma (a weaker version) for locally compact spaces on Walter Rudin's Real and Complex Analysis

The problem is that the Urysohn Lemma is valid for Normal Spaces and we are trying to prove something similar, in fact weaker, but for locally compact spaces. Recall that a locally compact space is completely regular, but there are locally compact spaces that are not normal (I am aware of the existence of such spaces, although a counterexample does not come to mind) and hence the full force of the Urysohn's Lemma does not apply.

Still, for what you are looking for, Rudin has it all covered up excellently and I couldn't do it in a more concise way here other than copy and pasting it, so you'd better look it upon pages 35 40 of the aforementioned book.

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  • $\begingroup$ Thank you very much for the source! :) $\endgroup$ – Yaddle Jun 10 '16 at 11:49
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    $\begingroup$ The Tychonoff plank is a standard example for what you stated. It's $((\omega_1 + 1) \times (\omega+1)) \setminus \{(\omega_1,\omega)\}$, where the ordinal numbers have the order topology etc. $\endgroup$ – Henno Brandsma Jun 10 '16 at 12:08
  • $\begingroup$ You're welcome. And @HennoBrandsma thank you for the counterexample, I didn't know that. I'm going to check it out. $\endgroup$ – proofromthebook Jun 10 '16 at 12:33
  • $\begingroup$ See en.wikipedia.org/wiki/Tychonoff_plank, what I described is there called the deleted Tychonoff plank, BTW. $\endgroup$ – Henno Brandsma Jun 10 '16 at 12:46
  • $\begingroup$ And topology.jdabbs.com/… will query a database of spaces for such examples, and it has 4 (one of which was mine). $\endgroup$ – Henno Brandsma Jun 10 '16 at 12:48

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