Proving a subsequence converges to S

**hockey777** · September 24th 2008, 12:32 PM

Let $(x_n)$ be a bounded sequence for each $n \in N$ let $s_n:=$ sup{ $x_k:k$ > $n$ } and $S=inf$ { $s_n$ }. Show that there exists a sub sequence of $(x_n)$ that converges to S.

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

Since it is decreasing and bounded then $s_n$ converges to $S$ . Futhermore there exists a subsequence of $s_n$ that is convergent that much also converge to S. My only problem is I don't know how to relate this back to $(x_n)$

**hockey777** · September 24th 2008, 12:51 PM

I take that back, there is still need for help.

**ThePerfectHacker** · September 24th 2008, 12:56 PM

Originally Posted by hockey777

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

This does not make sense, $s_n$ are sets, how can they be "larger" or "smaller"?

**hockey777** · September 24th 2008, 01:00 PM

Originally Posted by ThePerfectHacker

This does not make sense, $s_n$ are sets, how can they be "larger" or "smaller"?

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

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