Characteristic property of a $T_1$ topological space Hello I have problems with this exercise
A topological space is called $T_1$ if every finite subset is closed.
Prove that $X$ is a $T_1$ space if and only if $\{x \} = \displaystyle\bigcap_{U \in  I(x)}^{}{U}$ for every $x \in X$
EDIT: Let’s denote by $I(x)$ the family of neighbourhoods of $x$.
Thanks
 A: This is true, since it holds if and only if there's a nbhd of $x, \quad U_y $ such that  $y\notin U_y$, for each $y\ne x$.
(It's also equivalent to  $\{x\} $ being closed,  for all $x $.)
A: Suppose $X$ is $T_1$. We clearly have that $x \in O$ for every (open) neighbourhood of $x$ (so for all $O \in \mathcal{T}(x)$). So $$\{x\} \subseteq \bigcap_{O \in \mathcal{T}(x)} O$$ is immediate and for free. Now let $y \in \bigcap_{O \in \mathcal{T}(x)} O$, and we want to see that $y \in \{x\}$ or $y=x$. So assume (for a contradiction) that $y \neq x$. We note that $\{y\}$ is closed by the $T_1$-ness of $X$, and also $x \in X\setminus\{y\}$, so that $X\setminus\{y\} \in \mathcal{T}(x)$ and it follows that $$y \in \bigcap_{O \in \mathcal{T}(x)} O \subseteq X\setminus\{y\}$$ as the intersection is a subset of any of its intersecting sets. But this is an immediate contradiction so that $y \neq x$ cannot hold. So $$\bigcap_{O \in \mathcal{T}(x)} O \subseteq \{x\}$$ and we have shown the identity.
The reverse implication is quite similar, and just a hint: to show all finite sets to be closed, it is enough to show that all singletons are.
