Subsequence of a bounded sequence converges to a standard part of the hyperreal This is a homework problem.
If $(x_n)$ is a bounded sequence of real numbers, prove that there exists a subsequence of $(x_n)$ that converges towards the standard part of the hyperreal $[(x_n)]$, ie. the equivalence class of $(x_n)$.
Now a standard part of a hyperreal $[(x_n)]$ is a number $\alpha$ such that $ \forall r>0, \ \ |(x_n-\alpha)|<r$ for almost-all $n$, or $\mathscr{U}$-almost-all, $\mathscr{U}$ is the ultrafilter.
Now if a subsequence $(x_k)$ converges to $\alpha$, then $\forall \epsilon>0 \ \ \exists N_{\epsilon} \in  \mathbb{N}, \ \forall k>N_{\epsilon},\ \ |x_k-\alpha|<\epsilon $
Now I would need to use the fact that $ \forall r>0, \ \ |(x_n-\alpha)|<r$ for $\mathscr{U}$-almost-all $n$.
The problem I have is basically that it's been far too long since I last needed this kind of proper math, sequences and so, have been to simulation and probability theory and such for too long.

I see that the question was answered via "comments".  It may be instructive to mention that it may not be possible to represent the hyperreal $[(x_n)]$ as the class of a convergent subsequence (or any other sequence).  Namely, the subsequence constructed above may be supported on a "small" set of indices (i.e. not in the ultrafilter).  However, if the ultrafilter is a P-point, this will be possible.  
Another way of explaining the construction of the subsequence is to note that the real st$([(x_n)])$ must be an accumulation point: otherwise $x_n$ can be bounded away from it by a positive $\epsilon$.
