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Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:

Prove that a topological space $X$ is $\mathrm{T2}\iff\{x\}=\bigcap\limits_{U\in\,\mathcal{I}(x)}\kern{-7pt}\overline{U}\;\;\;\forall x\in X$

Partial proof: $(\Rightarrow)$ Let $x\in X$, suppose that $y\in X$ and $x\neq y$. They have disjoint neighborhoods because $X$ is $\mathrm{T2}$. So $\exists A,B\subset X$ open and disjoint, such that $x\in A$ and $y\in B$, and as they're disjoint $A\subset B^{\kern{1pt}\mathrm{c}}$ and $B^{\kern{1pt}\mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $y\neq x$ yields a family of closed neighborhoods of $x$, and $\bigcap\limits_{y\neq x}\kern{-3pt}\overline{B^{\kern{1pt}\mathrm{c}}_y}=\bigcap\limits_{y\neq x}\kern{-3pt}B^{\kern{1pt}\mathrm{c}}_y=\{x\}$.

$(\Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,y\in X$ they have no $A\in\mathcal{I}(x),B\in\mathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?

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Suppose that $X$ is $T_2$. If $x,y\in X$, there are $U\in\mathcal{I}(x)$ and $V\in\mathcal{I}(Y)$ such that $U\cap V=\emptyset$. Let $O$ be an open set such that $y\in O$ and $O\subset V$. Then $O^\complement$ is a closed set that containes $U$. Therefore, $\overline U\subset O^\complement$ and therefore $y\notin\overline U$. Sinse this occurs for every $y\in X$, $\bigcap_{U\in\mathcal{I}(x)}\overline{U}=\{x\}$.

Now, suppose that $X$ is not $T_2$. Take $x,y\in X$ such that, for any $U\in\mathcal{I}(x)$ and any $V\in\mathcal{I}(y)$, $U\cap V\neq\emptyset$. It's not hard to prove that $y\in\bigcap_{U\in\mathcal{I}(x)}\overline{U}$.

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  • $\begingroup$ Thanks for the suggestion! $\endgroup$ – Ladooscuro Dec 27 '18 at 20:35
  • $\begingroup$ Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B" $\endgroup$ – Ladooscuro Dec 27 '18 at 20:45
  • $\begingroup$ Indeed it does. $\endgroup$ – José Carlos Santos Dec 27 '18 at 20:45
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Starting the proof with “let $x,y\in X$” is not the right way.

Suppose $X$ is Hausdorff and let $x\in X$. Suppose $y\in X$ and $y\ne x$. Then there exist $U\in\mathcal{I}(x)$ and $V\in\mathcal{I}(y)$ with $U\cap V=\emptyset$; in particular $y\notin\bar{U}$. Thus $y\notin\bigcap_{U\in\mathcal{I}(x)}\bar{U}$.

Suppose $X$ is not Hausdorff and let $x,y\in X$, with $x\ne y$, such that $U\cap V\ne\emptyset$, for every $U\in\mathcal{I}(x)$ and $V\in\mathcal{I}(y)$. In particular, $y\in\bar{U}$, for every $U\in\mathcal{I}(x)$.

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  • $\begingroup$ Care to explain why that's not the right way? except for not assuming x \neq y $\endgroup$ – Ladooscuro Dec 27 '18 at 21:24
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    $\begingroup$ @Ladooscuro Well, that's a very important point. Also you want to show something about any $x\in X$. So it's better to start with $x$ and introduce the auxiliary $y$ later. $\endgroup$ – egreg Dec 27 '18 at 21:36
  • $\begingroup$ Thanks, i'll keep it in mind $\endgroup$ – Ladooscuro Dec 27 '18 at 21:37

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