Property of Hausdorff spaces I want to show that $$\text{E is a Hausdorff space} \iff \bigcap_{V(x) \text{ is a closed neighborhood of X}} V(x)=\{x\}$$
I am sure my proof is incorrect, since along the way, I have managed to proove that $$\text{E is a Hausdorff space} \iff \bigcap_{V(x) \text{is a neighborhood of X}} V(x)=\{x\}$$ and it must be a $\implies$ only...
Here is what I did :
$\begin{aligned}
\text{E is not a Hausdorff space} & \iff \forall V \in V(x) \quad \forall U\in V(Y), \quad U\cap V \neq \varnothing \\
& \iff \exists z\in E : z\in U\cap V \quad \forall V \in V(x) \quad \forall U\in V(Y) \\
& \iff \exists z\in E : z\in V \quad \forall V \in V(x) \\
& \iff \exists z\in E : z\in \bigcap V \quad \text{where V is a neighborhood of }x \\
& \iff \exists z\in E : z\in \bigcap V_F \quad \text{where $V_F$ is a closed neighborhood of }x \\
& \iff \text{the intersection of all neighborhoods of }x \\
& \phantom{\iff} \ \text{ 
does contain another element than }x 
\end{aligned}$
Which proves the statement.
Where did I make a mistake ?
 A: You’ve badly mismanaged your quantifiers. I strongly suggest that you write out your proofs in normal mathematical prose: it’s generally more readable, and on the whole you’re more likely to spot problems.
The first error isn’t catastrophic, but it is an error: the statement that $E$ is not Hausdorff means that there exist points $x$ and $y$ such that $x\ne y$, and $U\cap V\ne\varnothing$ for each $U\in V(x)$ and $V\in V(y)$. The next step, though, is fatal: you’re claiming that there is a single point $z$ that is in $U\cap V$ no matter which $U\in V(x)$ and $V\in V(y)$ we choose, and there’s no reason to think that this is the case. What you mean is that for each $U\in V(x)$ and $V\in V(y)$ there is a point $z(U,V)\in U\cap V$; the indexing makes it clear that this point depends on the choice of $U$ and $V$.
A better idea is to prove the result directly. Fix $x\in E$. If $y\in E\setminus\{x\}$, there are $U_y\in V(x)$ and $V_y\in V(y)$ such that $U_y\cap V_y=\varnothing$. Thus, $V_y$ is an open nbhd of $y$ disjoint from $U_y$, so $y\notin\operatorname{cl}U_y$. This means that $\operatorname{cl}U_y$ is a closed nbhd of $x$ that does not contain $y$, and at this point you’re practically done.
A: Taking different $U$ and $V$ may give you a different $z$. By assuming that the $z$ is the same, you assume the wanted conclusion.
A: the Hausdorff condition that distinct points have distinct open neighborhoods implies that a a point $x$ has a closed neighborhood disjoint from any point $y \ne x$. since $x$ belongs to all its closed neighborhoods, the conclusion follows
