If $U$ is an ultrafilter on $\mathbb{N}$, then $U$ limits exist. This is rather silly, I expect Asaf will point out what I am missing immediately.
Let $U$ be a filter on $\mathbb{N}$. If $\{a_n\}_{n=1}^\infty$ is a sequence of reals, we write $\lim_U a_n = a$ if for each $\epsilon > 0$ 
$$
\{ n \ : \ |a_n - a|< \epsilon \} \in U
$$
I want to show that if $U$ is an ultrafilter, then $U$ limits exist for every bounded sequence of reals $\{a_n\}$. Showing they are unique is easy. Also, this is easy if $U$ is principle (if it's generated by $n$, then $a_n$ is the limit). Additionally, I have proved the result when the sequence is convergent (As, assuming $U$ is non-principle, $U$ contains all the co-finite sets).
Haven't gotten much further after playing around. I tried breaking $a_n$ into disjoint convergent subsequences (you can do this but there may be $\omega$ many). But the lack of unity in my approach makes me think I am going about this the wrong way. Anyone have a proof?
 A: Let $I_0$ be a closed interval containing the sequence. Subivide $I_0$ in half, and for one of those halves, a closed interval denoted $I_1$, some subsequence is in $I_1$ on a set of the ultrafilter. Subdivide $I_1$ in half, and for one of those halves, a closed interval denoted $I_2$, some subsequence is in $I_2$ on a set of the ultrafilter. Continue inductively. The intersection $I_0 \cap I_1 \cap I_2 \cap I_3 \cap \cdots$ is a single point $a$. And $\lim_U a_n = a$ is pretty easy to see now.
A: Let $\langle a_k:k\in\Bbb N\rangle$ be a bounded sequence in $\Bbb R$, and let $M$ be such that $|a_k|\le M$ for each $k\in\Bbb N$. Suppose that for each $x\in[-M,M]$ there is an $m(x)\in\Bbb N$ such that 
$$\left\{k\in\Bbb N:|a_k-x|<2^{-m(x)}\right\}\notin U\;.$$
For each $x\in[-M,M]$ let $B_x=\left\{y\in\Bbb R:|x-y|<2^{-m(x)}\right\}$, and let $\mathscr{B}=\{B_x:x\in[-M,M]\}$; clearly $\mathscr{B}$ is an open cover of the compact set $[-M,M]$, so there is a finite $F\subseteq[-M,M]$ such that
$$[-M,M]\subseteq\bigcup_{x\in F}B_k\;.$$
$U$ is an ultrafilter, and $F$ is finite, so there must be an $x\in F$ such that $\{k\in\Bbb N:a_k\in B_x\}\in U$, contradicting the choice of $m(x)$.
