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.

  • $\begingroup$ You need to show that all co-finte subsets of $\mathbb N$ are in $\mathcal U$. By definition of "ultrafilter" it suffices to show that $\mathcal U$ has no finite sets as elements, which again is a consequence of the ultrafilter not being trivial. $\endgroup$ – Hagen von Eitzen Nov 22 '12 at 12:38
  • $\begingroup$ If I select a number $r$, is the cardinality of the set $\{ k \in \mathbb{N} \ | \ \ |x_k-\alpha| > r \} < \infty$? $\endgroup$ – Valtteri Nov 22 '12 at 12:57
  • $\begingroup$ Yes, because with $\epsilon=r>0$, we can have $x_k-\alpha|>r$ only for $k$ with $k\le N_\epsilon$. $\endgroup$ – Hagen von Eitzen Nov 22 '12 at 13:02
  • $\begingroup$ OK, here's my thoughts: If we select $\epsilon_1>0$ then there exists the smallest $k$ such that $k \in \mathbb{N}, \ \ |x_k-\alpha|<\epsilon_1$. Then we set $\epsilon_m = \frac{\epsilon_{m-1}}{2}$. Now we form a sequence of smallest ks for which $k \in \mathbb{N}, \ \ |x_k-\alpha|<\epsilon_m$. Thus we get a subsequence of $x_n$, namely $x_{k_{min}}$ and it converges to $\alpha$. $\endgroup$ – Valtteri Nov 22 '12 at 13:13
  • $\begingroup$ Yes, that works. $\endgroup$ – Brian M. Scott Nov 22 '12 at 13:27

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 $\epsilon$.


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