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Let $c$ be a cluster point of a bounded sequence $(x_n)$. I am trying to prove that there exists a free filter $\mathcal{F}$ s.t.

$$c= \lim_\mathcal{F} \ (x_n) .$$

My attempt: Since $c$ is a cluster point of $(x_n)$, for every neighbourhood $V$ of $c$ and every $k \in \mathbb{N}$,

$$\{n \geq k : x_n \in V \} \neq \emptyset$$

It is enough to define

$$ \mathcal{F}:= \big\{ \{n \geq k : x_n \in V \} \big\}_{k \in \mathbb{N}, \ V \in \mathcal{V}(c)}$$

because it is a free filter and $c= \lim_\mathcal{F} \ (x_n): $

Free filter:

1.- Since $\{n \geq k : x_n \in V \} \neq \emptyset$, $\emptyset \notin \mathcal{F}$.

2.- $\{n \geq k_1 : x_n \in V_1 \} \cap \{n \geq k_2 : x_n \in V_2 \} = \{n \geq \max \{ k_1,k_2\} : x_n \in V_1 \cap V_2 \} \neq \emptyset $

3.- Is every $A$ s.t. $\{n \geq k : x_n \in V \} \subset A \subset \mathbb{N}$ also in $\mathcal{F}$? I am not sure of it, maybe not but $\mathcal{F}$ is basis... .

4.- $ \cap \mathcal{F} = \emptyset$ as $k \in \mathbb{N}$. No $n$ can be in every set!

For $c= \lim_\mathcal{F} \ (x_n) $ it is enough to take $k=1$, so $\{n : x_n \in V \} \in \mathcal{F}$ $\blacksquare$

Question is: is this proof correct? What about 3.- ?

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    You need to give complete proof of the fact that $\mathcal{F}$ is a filter (or a filterbase, so generates a filter) an the limit assertion also needs a small proof. Also, why is the filter free? – 2017-01-05
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    @HennoBrandsma I have just written the details. Although at first I thought every condition was right, I am not sure of #3. Is that true or false in general? If it is false, is $\mathcal{F}$ basis of a filter? – 2017-01-05

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Corrected. Let $C=\{n\in\Bbb N:x_n=c\}$. Your approach works fine if $C=\varnothing$, and you can prove $(3)$ as follows.

For each $V\in\mathscr{V}(c)$ and $k\in\Bbb N$ let $F_V(k)=\{n\ge k:x_n\in V\}$. Suppose that $A\supseteq F_V(k)\in\mathscr{F}$. Let $U=V\cup\{x_n:n\in A\}$; then $U\supseteq V\in\mathscr{V}$, so $U\in\mathscr{V}$, and $A=F_U(0)\in\mathscr{F}$.

If $C$ is infinite, let $\mathscr{F}=\{F\subseteq\Bbb N:C\setminus F\text{ is finite}\}$; it’s not hard to show that $\mathscr{F}$ is a free filter on $\Bbb N$ such that the $\mathscr{F}$-limit of $\langle x_n:n\in\Bbb N\rangle$ is $c$, since $\{n\in\Bbb N:x_n\in V\}\supseteq C$ for each $V\in\mathscr{V}$.

If $C$ is finite but non-empty, it seems easiest first to construct a filter on $\Bbb N\setminus C$ as in the second paragraph: let $M=\Bbb N\setminus C$, for each $V\in\mathscr{V}(c)$ and $k\in\Bbb N$ let

$$F_V(k)=\{n\in M:n\ge k\text{ and }x_n\in V\}\;,$$

and let $\mathscr{F}=\{F_V(k):k\in\Bbb N\text{ and }V\in\mathscr{V}\}$. Then, as in the case $C=\varnothing$, $\mathscr{F}$ is a free filter on $M$, and $c$ is the $\mathscr{F}$-limit of $\langle x_n:n\in M\rangle$. $\mathscr{F}$ is a base for the free filter

$$\mathscr{G}=\{F\cup S:F\in\mathscr{F}\text{ and }S\subseteq C\}$$

on $\Bbb N$, and $c$ is the $\mathscr{G}$-limit of $\langle x_n:n\in\Bbb N\rangle$.

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    I'm afraid I don't understand #3 at all. Since $N$ is a neighbourhood of $c$, it contains an open set $(c-\varepsilon,c+\varepsilon)$. So if you claim that $N \subset M$, then we have $(c-\varepsilon,c+\varepsilon) \subset M$. But $M$ has a countable amount of points, since $A \subset \mathbb{N}$! I don't understand this. – 2017-01-06
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    @Minkowski: Sorry: that was a typo. Fixed now; thanks for catching it. – 2017-01-06
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    It would be also extremelly helpful if you could tell me the difference between the set $\mathcal{F}$ I defined (and we agreed it was basis of a filter) and the set $\mathscr{F}$ you defined. I don't really get the difference between them. They seem the same to me! – 2017-01-06
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    @Minkowski: And I have to apologize again: somehow I misread *neighbourhood* as *open neighbourhood*. Your version is fine, except that you need to prove $(3)$. I’ll rewrite my answer to reflect that. – 2017-01-06
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    @Minkowski: I also removed the comment about the typo, since it no longer applies. – 2017-01-06
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    It is perfectly clear now for me. Thanks and apologies for the bother (you wrote a clear answer and you had to remove it). – 2017-01-06
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    @Minkowski: No problem; I apologize again for misreading your original attempt! – 2017-01-06
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    Sorry for bothering you again Brian, but reading again the answer I think either I don't understand something or there must be another typo in the answer: given $V \in \mathcal{V}(c)$ and $k>0$ (big) consider a sequence that $x_0 \in V$. Then we choose $F_V(k)= \{n \geq k: x_n \in V \}$. For sure $0 \not\in F_V(k)$. Now, if $F_V(k) \subseteq A$ but $1 \not\in A$, then we will have that $0 \in F_U(0)$ because $V \subseteq U$, but $0 \not\in A$ ! – 2017-01-10
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    I will give you an example of what I mean because I'm not sure my last message was clear (I will count from 0 as you wrote $F_V(0)$): consider the secuence (5, -1,-2,-3,-4, 6, 5.5, 5.33, 5.25, 5.20, .....). Then $F_V(3):=\{n \geq 3 : x_n \in V=(4.8, 5.2) \}=\{10,11,12,...\} $ . Now let $A= \{8,9,10,11,...\}$. Then $U=(4.8,5.2] \cup \{ 5.25 \}$. But then $0 \in F_U(0)$ and $0 \not\in A$! Am I making myself clear? – 2017-01-10
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    **In my first comment** it should be "...but $0 \not\in A$, then we will...", sorry about that. – 2017-01-10
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    @Minkowski: Sorry to be so slow; I was mostly offline yesterday. That’s a genuine oversight on my part; let me think about it for a while to see whether I can make a simple correction. – 2017-01-11
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    @Minkowski: And I’ve now corrected the answer to cover the messier cases in which $c$ is one or more terms of the sequence. – 2017-01-12
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    Thank you very much Brian. I see it totally clear now. – 2017-01-12
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    @Minkowski: You're very welcome. – 2017-01-12