Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Hagen von Eitzen Nov 22 '12 at 12:38
    
If I select a number $r$, is the cardinality of the set $\{ k \in \mathbb{N} \ | \ \ |x_k-\alpha| > r \} < \infty$? –  Valtteri Nov 22 '12 at 12:57
    
Yes, because with $\epsilon=r>0$, we can have $x_k-\alpha|>r$ only for $k$ with $k\le N_\epsilon$. –  Hagen von Eitzen Nov 22 '12 at 13:02
    
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 $\alpha$. –  Valtteri Nov 22 '12 at 13:13
    
Yes, that works. –  Brian M. Scott Nov 22 '12 at 13:27

1 Answer 1

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 $\epsilon$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.