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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).2012-11-17
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    thank you for this clarification, but am asking about the proof that i wrote above if it's correct ?2012-11-17
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    If you want people to be able to judge your proof first you need to make it clear. Confusing the notations will confuse readers, and some readers would argue that your proof is therefore incorrect.2012-11-17
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    Ok, but i didn't know the difference at 1st.If you look to the end of the first line i wrote where $ F \in$ $\mathscr F$.2012-11-17
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    @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 $$x\in\bigcap_{F\in\mathscr{F}}\overline F\;.$$2012-11-17
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    Your argument is basically correct. There are places where you could say things more clearly or more accurately; besides the one already mentioned, you should say that $\mathscr{U}_x$ is the nbhd system **at** $x$. However, it’s clear that you understand why the theorem is true.2012-11-17
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    @ Brian M. Scott its my second time with MathJax basic i should write it as u say .2012-11-17
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    If you’ve not already seen it, you might find [this MathJax link](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) helpful.2012-11-17
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    @ Brian M. Scott could u tell me how u wrote the second one so i can edit it?2012-11-17
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    LaTeX hint: there is no need to add `$` between symbols.2012-11-17
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    Right-click on any MathJax formula, select *Show Math As* and then *TeX Commands*, and you’ll get a pop-up window showing the code that was used to produce the formula.2012-11-17
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    @BrianM.Scott ;@Asaf Karagila thanks.2012-11-17
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    @leo a filter $\mathscr F$ is a nonempty collection of subsets of $\mathbf X$ , where is the problem?2012-11-17
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    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.2012-11-18
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    @AndreasBlass thank you i will delete them from the proof. ;@leo Let $\mathbf E \subset \mathbf X$ , a point $x \in \mathbf X$ is a cluster point at $\mathbf E$ iff every nbd of x contains a point of $\mathbf E$ different from x.2012-11-18

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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.