The theorem reads: If X is a metric space and E is a subset of X, then the closure of E is closed.(adsbygoogle = window.adsbygoogle || []).push({});

Rudin's proof: If p belongs to X and not to the closure of E, then p is neither a point of E nor a limit point of E. Hence p has a neighborhood which does not intersect E.The complement of the closure of E is therefore open.Hence the closure of E is closed (POMA, 35).

I'm having trouble seeing why the italicized portion of the proof is correct. Yes, p has a neighborhood which does not intersect E, but, (if I'm not mistaken) the statement that the complement of the closure of E is open does not follow from the fact that p has such a neighborhood. p, it seems, would need to have a neighborhood which does not intersect the closure of E (and not just E itself), for a neighborhood of p that contains limit points of E but not any points of E itself would not be a subset of the complement of the closure of E (in which case, p would not be an interior point of the complement of the closure of E).

Now, intuitively, it seems that if p has a neighborhood which does not intersect E, it cannot have one that intersects the set of limit points of E. However, the fact that Rudin does not mention this statement suggests that there is something about his proof that I do not understand.

Thanks in advance for any help!

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# Part (a) of Theorem 2.27 in Rudin's POMA

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