Filters on topology Show that the filter $\mathscr F$ has $x$ as a cluster point iff $x \in\bigcap_{F \in \mathscr F } \overline  F$.
For the the Proof of the 1st direction $(\Rightarrow)$ :
Let the filter $\mathscr F$ has $x$ as a cluster point so every element $F$ of $\mathscr F$  intersect every $U \in $ $\mathscr U_x$ where  $\mathscr U_x$ is the nbd system ,since every $U$ intersects $F$ , so $x \in\overline F \Rightarrow x \in \cap \overline F$.
Now for the other direction $(\Leftarrow)$ :
Let $x \in \cap \overline  F$  so this mean that $x$ belongs to each $\overline  F$ then every nbd $U$ of $x$ intersect $F$, and $F$ is elements of  $\mathscr F$, then $x$ is a cluster point of $\mathscr F$.
If any one can tell me that my proof is correct or not ?
 A: Write it down more calmly, using more sentences etc.:
Suppose $x$ is a cluster point of $\mathscr F$. We want to show that $x \in\bigcap_{F \in \mathscr F } \overline  F$, and so pick an arbitrary $F \in \mathscr F$. To see $x \in \overline F$, we pick any open neighbourhood $O$ of $x$, and we need to see that $O$ intersects $F$. But this is clear from the definition of $x$ being a cluster point got $\mathscr F$. The reverse is similar.
If you want to see it more as a logical fact plus definitions:
$x$ is a cluster point of $\mathscr F$ means by definition $\forall O \in \mathscr{U}_{x} : \forall F \in \mathscr{F}: O \cap F \neq \emptyset$ which is the same as 
$\forall F \in \mathscr{F}: ( \forall O \in \mathscr{U}_{x} : O \cap F \neq \emptyset)$, and the statement in brackets is by (a) definition the meaning of $x \in \overline{F}$, so it also says $ \forall F \in \mathscr{F}: x \in \overline{F} $, which by definition of intersection is
just $x \in \bigcap_{F \in \mathscr F} \overline{F} $.
So the fact is just a simple restatement of the definions of closure, intersection and cluster point.
