Convergence of real numbers along an ultrafilter Suppose we have an uncountable set $I$ and a non-principal ultrafilter $U$ on it. I am interested whether it is possible to conclude that if cardinality of $I$ is big enough, then every tuple $(x_i)_{i\in I}$ which converges to 0 along $U$ is actually constantly null on some set from $U$. 
I suspect that for this to happen the cardinality of $I$ must be huge as seemingly there is no such set for $|I|=\aleph_\omega$. Is it really possible?
 A: Let us assume to avoid trivialities that $U$ is required to contain no sets of cardinality smaller than $|I|$.  Even with this assumption, no matter how large $I$ is, your desired conclusion will not hold for arbitrary ultrafilters.  Indeed, let $I$ be any infinite set and partition $I$ into countably many sets $I_n$ with the same cardinality as $I$.  Define $(x_i)_{i\in I}$ by $x_i=1/n$ if $i\in I_n$.  Now let $F$ be the filter generated by the sets $J_N=\bigcup_{n>N} I_n$ for each $N$ and all the sets whose complements have cardinality $<|I|$.  This is clearly a proper filter, so extend it to an ultrafilter $U$.  The ultrafilter $U$ now contains no sets of cardinality $<|I|$, and $(x_i)$ converges to $0$ along $U$ since $J_N\in U$ for all $N$.  However, there is not even a single value of $x_i$ which is $0$.
However, if you put much stronger conditions on $U$ then you can get this conclusion.  In particular, if you assume $U$ is closed under countable intersections, then if $(x_i)$ converges to $0$ along $U$ then almost every $x_i$ must be $0$.  Indeed, for each $n$ the set $A_n=\{i:|x_i|<1/n\}$ is in $U$, and hence $\bigcap A_n=\{i:x_i=0\}$ is also in $U$.  (In fact, conversely, it is not hard to show by a construction similar to the previous paragraph that if $U$ is not closed under countable intersections, then there is a tuple $(x_i)$ that converges to $0$ along $U$ with $x_i\neq 0$ for all $i$.)
However, ultrafilters that are closed under countable intersections are rather difficult to find!  By definition, for such an ultrafilter $U$ to exist on a set $I$, $I$ must be at least as large as the smallest measurable cardinal.  Measurable cardinals cannot be proven to exist in ZFC, and are a fairly strong kind of "large cardinal".
