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I studied from Munkres' Topology its proof. He says that the Hausdorff condition enables us to choose, for each $b \in B$ an open set $U_b$ whose closure is disjoint from $a$. I did not get why closure does not intersect $a$ as Hausdorff only says that we can find disjoint open sets or neighbourhoods for each point of a Hausdorff space? Either it is trivial for closed set also?

Henno Brandsma
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Hafsa
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1 Answers1

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Let's call our space $X.$ Since $X$ is Hausdorf, then for distinct $a,b\in X,$ there are open sets $U_a$ and $U_b$ with $a\in A_u,$ $b\in U_b,$ and $U_a\cap U_b=\emptyset.$ Thus, $U_b\subseteq X\setminus U_a.$ Since $U_a$ is open, then $X\setminus U_a$ is closed. The closure of $U_b$ is the intersection of all closed supersets of $U_b,$ so the closure of $U_b$ is a subset of $X\setminus U_a,$ so that $U_a$ is disjoint from the closure of $U_b.$ In particular, $a\notin \overline{U_b}.$

Henno Brandsma
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Cameron Buie
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  • I have an other question that why we are prving that clouser of open set is disjoint from a? is it not enough to show that Ub is disjoint from a? – Hafsa Dec 13 '19 at 09:02
  • Honestly, I'm not sure, and I don't have the leisure to think about it. If it were enough, though, then we'd instead be able to prove the stronger result that every paracompact $T_1$ space is normal. If your suspicion is correct, then that result is true. If that result is false, then it isn't enough that the sets $U_b$ are disjoint from $a.$ You can try asking that as another question here, or dig into how the proof uses paracompactness to see where the disjoint closures might be important. – Cameron Buie Dec 13 '19 at 12:20
  • @Hafsa Look at my explanation at the other version of this question and see where it is used. – Henno Brandsma Dec 14 '19 at 14:55
  • I asked an other question and some one told me. – Hafsa Dec 14 '19 at 15:09