Can you please help me with this question?
Let's $X$ be a topological space. Show that these two following conditions are equivalent :
$X$ is hausdorff
for all $x\in X$ intersection of all closed sets containing the neighborhoods of $x$ it's $\{x\}$.
Thanks a lot!
HINTS:
If $x$ and $y$ are distinct points in a Hausdorff space, they have disjoint open nbhds $V_x$ and $V_y$, and $X\setminus V_y$ is a closed set containing $V_x$.
If $F$ is a closed set containing an open nbhd $V$ of $x$, then $V$ and $X\setminus F$ are disjoint open sets.
Recall the definition of Hausdorff:
$X$ is a Hausdorff space if for every two distinct $x,y\in X$ there are disjoint open sets $U,V$ such that $x\in U$ and $y\in V$.
Suppose that $X$ is Hausdorff and $x\in X$. Suppose $y\neq x$. We have $U,V$ as in the definition. So $x\in U$ and $y\in V$ and $U\cap V=\varnothing$. Suppose that $F$ is a closed set containing an open neighborhood of $x$, intersecting this open set with $U$ yields an open neighborhood of $x$ which is a subset of $F$ so without loss of generality $U\subseteq F$. $F'=F\cap(X\setminus V)$ is closed and does not contain $y$. Furthermore $U$ itself is a subset of this closed set and $y\notin F'$. Therefore when intersecting all closed subsets which contain an open environment of $x$ we remove every other $y$, so the result is $\{x\}$.
On the other hand, suppose that for every $x\in X$ this intersection is $\{x\}$, by a similar process as above deduce that $X$ is Hausdorff as in the definition above.
