Let $\omega$ be a [nonprincipal ultrafilter](http://en.wikipedia.org/wiki/Ultrafilter) on $\mathbb N$.


A standard construction gives an $\omega$-limit, say $x\_\omega$, for any bounded sequence $(x\_n)$ of real numbers. 
Namely, there is unique real value $x\_\omega$ such that  

$$\{\,n\in\mathbb N\mid |x\_\omega-x\_n|<\varepsilon\,\}\in \omega$$
for any $\varepsilon>0$.


Clearly $x\_\omega$ is a partial limit of $x\_n$ [i.e., $x\_\omega$ is a limit of a subsequence of $(x\_n)$].



> 
> **Question.** Is it always possible to choose subsequence $(x\_n)$, $n\in J$
>  converging to $x\_\omega$ and such that $J\in\omega$?
> 
> 
> 



Ultralimit versus partial limit
No, only if $\omega$ is a p-point. 


If $(A\_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A\_k$ in a finite set, then we can choose a sequence $x\_n$ by declaring $x\_n:= 1/k$ whenever $n\in A\_k$. On each set $J\in \omega$ the sequence $x\_n$ does not converge to 0. 


To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A\_k$, let $I$ be the ideal of sets meeting each $A\_k$ only finitely often (except for finitely many $k$), and let $\omega$ be dual to any maximal ideal extending $I$. 
Then each $A\_k$ is in $I$, hence null modulo $\omega$. The partition $(A\_k)$ witnesses that $\omega$ is not a p-point. 


Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions). 


