# Proving a subsequence converges to S

• Sep 24th 2008, 12:32 PM
hockey777
Proving a subsequence converges to S
Let $\displaystyle (x_n)$ be a bounded sequence for each $\displaystyle n \in N$ let $\displaystyle s_n:=$sup{$\displaystyle x_k:k$>$\displaystyle n$} and $\displaystyle S=inf${$\displaystyle s_n$}. Show that there exists a sub sequence of $\displaystyle (x_n)$ that converges to S.

Ok so clearly $\displaystyle s_1$>$\displaystyle s_2$>....>$\displaystyle s_n$.

Since it is decreasing and bounded then $\displaystyle s_n$ converges to $\displaystyle S$. Futhermore there exists a subsequence of $\displaystyle s_n$ that is convergent that much also converge to S. My only problem is I don't know how to relate this back to $\displaystyle (x_n)$
• Sep 24th 2008, 12:51 PM
hockey777
I take that back, there is still need for help.
• Sep 24th 2008, 12:56 PM
ThePerfectHacker
Quote:

Originally Posted by hockey777
Ok so clearly $\displaystyle s_1$>$\displaystyle s_2$>....>$\displaystyle s_n$.

This does not make sense, $\displaystyle s_n$ are sets, how can they be "larger" or "smaller"?
• Sep 24th 2008, 01:00 PM
hockey777
Quote:

Originally Posted by ThePerfectHacker
This does not make sense, $\displaystyle s_n$ are sets, how can they be "larger" or "smaller"?

no $\displaystyle s_1$ is the first position in the sequence $\displaystyle s_n$, but I see where you think that as I messed up on my original post, I fixed it.