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Yet another question as part of this whole mess. Context: I'm trying to show that for a compact Hausdorff space X, the collection of sets $U_f=\{x\in X:f(x)\ne 0\}$ form a basis, where $f$ ranges over the continuous real valued functions on $X$. I've shown that this is A basis, now I'm trying to show that the topology this basis generates is the same topology as X.

I think it is sufficient to show that for any open set $U\in X$, and any point $x\in U$, $\exists f\in C(X)$ such that $x\in U_f\subset U$, because then any open set can be written as the union of these open basic sets.

So, if that's the case, I need to have a continuous function whose support is contained inside $U$, and that $x$ is an element of that support. This again smells like something like Urysohnn's lemma, but I'm not sure...

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  • $\begingroup$ It’s true for any Tikhonov space (i.e., completely regular Hausdorff space), and every compact Hausdorff space is Tikhonov; Uryson’s lemma is one way to prove that. $\endgroup$ – Brian M. Scott Dec 2 '15 at 8:15
  • $\begingroup$ @BrianM.Scott Thanks, it's been a year since I did general topology and I'm rusty, this is for a commutative algebra assignment. Pretty sure given the context that I can just quote the result :) $\endgroup$ – Alan Dec 2 '15 at 8:17
  • $\begingroup$ Yeah, that should be safe enough. :-) $\endgroup$ – Brian M. Scott Dec 2 '15 at 8:18

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