# A space is Hausdorff iff…

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?

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 $$y\in\bigcap_{U\in\mathcal{I}(x)}\overline{U}$$.

• Thanks for the suggestion! – Ladooscuro Dec 27 '18 at 20:35
• 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" – Ladooscuro Dec 27 '18 at 20:45
• Indeed it does. – José Carlos Santos Dec 27 '18 at 20:45

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 $$U\in\mathcal{I}(x)$$.

• Care to explain why that's not the right way? except for not assuming x \neq y – Ladooscuro Dec 27 '18 at 21:24
• @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. – egreg Dec 27 '18 at 21:36
• Thanks, i'll keep it in mind – Ladooscuro Dec 27 '18 at 21:37