# 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 ?

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Something is strange, I thought that filters contain sets, so $\bigcap\overline F$ is not well defined if $F\in\mathscr F$, because $\bigcap$ is defined on sets of sets. You might not want to skip on the notation $\bigcap\{\overline F\mid F\in\mathscr F\}$ (it takes effort to understand that this is what you mean, I think). – Asaf Karagila Nov 17 '12 at 22:12
@MissIndependent: Yes, you did, and that does make it possible to figure out what you meant. It would be clearer, however, if you wrote $$x\in\bigcap\{\overline F:F\in\mathscr{F}\}$$ or $$x\in\bigcap_{F\in\mathscr{F}}\overline F\;.$$ – Brian M. Scott Nov 17 '12 at 22:33
If you’ve not already seen it, you might find this MathJax link helpful. – Brian M. Scott Nov 17 '12 at 22:37
LaTeX hint: there is no need to add $ between symbols. – Asaf Karagila Nov 17 '12 at 22:46 The proposed argument seems to be a mixture of a correct proof with a couple of false and irrelevant statements. In the$\implies$direction, it need not be the case that$x\in F$, nor is this needed for the argument. In the converse direction, the claim that$\mathcal U_x\subseteq\mathcal F$and its consequence$\mathcal F\to x$need not be true, and again they're not actually needed. – Andreas Blass Nov 18 '12 at 3:47 ## 1 Answer 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.

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