Bounded non-convergent sequence with respect to an ultrafilter Let $\mathfrak{U}$ be an ultrafilter on $\mathbb{N}$ and $(x_n)$ be a real sequence. We say that $(x_n)$ is:


*

*bounded with respect to $\mathfrak{U}$, if there exists $M>0$ such that $\{n \mid |x_n|<M\} \in \mathfrak{U}$.

*convergent to $x$ with respect to $\mathfrak{U}$, if $\{n \mid |x_n-x|<\epsilon\} \in \mathfrak{U}$ for all $\epsilon>0$.
Are there a nonprincipal ultrafilter $\mathfrak{U}$ and a sequence $(x_n)$ such that $(x_n)$ is bounded but non-convergent with respect to $\mathfrak{U}$?
It probably exists, but any example of bounded sequence with respect to an ultrafilter that comes to my mind turns out to be convergent...
 A: There is no such ultrafilter. This is a consequence of the compactness of closed, bounded subsets of $\Bbb R$.
Let $\mathscr{U}$ be a non-principal ultrafilter on $\Bbb N$ and $\sigma=\langle x_n:n\in\Bbb N\rangle$ a sequence in $\Bbb R$ that is $\mathscr{U}$-bounded, and let $M\in\Bbb R$ be such that $\{n\in\Bbb N:|x_n|\le M\}\in\mathscr{U}$. If $|x|>M$, let $\epsilon=|x|-M$; then
$$\{n\in\Bbb N:|x_n-x|<\epsilon\}\subseteq\{n\in\Bbb N:|x_n|>M\}\notin\mathscr{U}\;,$$
so $\sigma$ clearly cannot converge to $x$ with respect to $\mathscr{U}$. Thus, any possible $\mathscr{U}$-limit of $\sigma$ must lie in the closed interval $[-M,M]$.
Suppose that $\sigma$ is not $\mathscr{U}$-convergent. Then for each $x\in[-M,M]$ there is an $\epsilon_x>0$ such that $\{n\in\Bbb N:|x_n-x|<\epsilon_x\}\notin\mathscr{U}$. For each $x\in[-M,M]$ let $B_x=(x-\epsilon_x,x+\epsilon_x)$; then $\{B_x:x\in[-M,M]\}$ is an open cover of the compact interval $[-M,M]$, so there is a finite $F\subseteq[-M,M]$ such that $$[-M,M]\subseteq\bigcup_{x\in F}B_x\;.$$
For each $x\in F$ let $N_x=\{n\in\Bbb N:x_n\in B_x\}$; by construction $N_x\notin\mathscr{U}$. Let $N=\bigcup_{x\in F}N_x$; $F$ is finite, so $N\notin\mathscr{U}$. But $\{n\in\Bbb N:|x_n|>M\}\notin\mathscr{U}$ as well, so 
$$\Bbb N=N\cup\{n\in\Bbb N:|x_n|>M\}\notin\mathscr{U}\;,$$
which is absurd. Thus, $\sigma$ must $\mathscr{U}$-converge to some $x\in[-M,M]$.
Added: By the way, this is an instance of a more general result. Let $X$ be a compact Hausdorff space, let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$, and let $\mathscr{U}$ be any ultrafilter on $\Bbb N$; then there is a (unique) $x\in X$ such that for each nbhd $V$ of $x$, $\{n\in\Bbb N:x_n\in V\}\in\mathscr{U}$. This $x$ is called the $\mathscr{U}$-limit of the sequence. (Note that if $\mathscr{U}$ is the principal ultrafilter at $m\in\Bbb N$, then the $\mathscr{U}$-limit of $\langle x_n:n\in\Bbb N\rangle$ is simply $x_m$.)
A: Let $U$ be a fixed ultrafilter. If $x_n$ is bounded (with respect to $U$), then $x_n$ is convergent (with respect to $U$).
If you know that generalised limits always exist in compact spaces, you can reason as follows. Since $x_n$ is bounded (wrt $U$), you can modify $x_n$ on for $n \in I$, $I \not \in U$ so that the resulting sequence, say $x'_n$, is bounded (with no additional quantification). It is simple to see that $x_n$ is convergent iff $x_n'$ is convergent. And $x_n'$ is convergent, because it's a sequence with elements in the compact space $[-M,M]$, and it converges to $U\!-\!\lim_n x'_n$, its generalised limit. 
To convince yourself that the generalised limit exists, you can proceed as follows. Start with a set $I_0 \in U$ such that for $n \in I$ you have $x_n \in A_0 := [-M,M]$. Next, once you have $I_k$ and $A_k$ defined, split $A_k = B \cup B'$, where $B,B'$ are halves of the interval $A_k$. Let $J = \{n \ | \ x_n \in B \}$ and $J'$ analogously. Either $J$ or $J'$ is in $U$, so take the one which is to be $I_{k+1}$, and take $A_{k+1} = B$ or $B'$ accordingly. Now, you can show without much work that $x_n$ converges to the sole point in $\bigcap_k A_k$ without too much trouble.
For more details (precise definitions, theorems) see Hindman-Strauss, especially section 3.5.
