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Maximal set with respect to the finite intersection property

Here, the accepted answer mentions (c) is false as stated. However, I don't understand why the following proof is incorrect:

Suppose $a,b$ are two points in $\cap D$ over each D in $\mathscr{D}$. Since X is T1, there is a nbd $U$ of $a$ not containing $b$. $a \in D$, hence $a \in \bar{D}$. By part a of the exercise, each nbd of $a$ lies in $\mathscr{D}$, which is a contradiction since $b$ is not in this nbd. Therefore there cannot be two points in the intersection.

HerrWarum
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    But it's the intersection of the closures, and $b$ can belong to the closure of each neighbourhood of $a$. – Daniel Fischer Nov 29 '17 at 21:29
  • @DanielFischer I see. Is this proof correct if the question is modified to say intersection over just the elements of $\mathscr{D}$ and not their closures? – HerrWarum Nov 29 '17 at 22:04
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    Yes, but in that case we don't need any conditions of the space, and a simpler argument shows that the intersection of a maximal family with the FIP is empty or a singleton. If $\bigcap D$ contains (at least) two points $a\neq b$, then we can adjoin $X\setminus {b}$ to the family and keep the FIP (even the total intersection remains nonempty), so it wasn't maximal. – Daniel Fischer Nov 29 '17 at 22:15
  • Got it, thanks. – HerrWarum Nov 29 '17 at 22:16

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