Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
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
1  
@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
1  
If you’ve not already seen it, you might find this MathJax link helpful. –  Brian M. Scott Nov 17 '12 at 22:37
1  
LaTeX hint: there is no need to add $ between symbols. –  Asaf Karagila Nov 17 '12 at 22:46
1  
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
show 10 more comments

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.