How would I go about proving such a statement? Any help would be great thanks.

- Oct 11th 2009, 01:14 PMdannyboycurtisAny bounded seq. has a converging subseq. <=> every bounded monotone seq. converges
How would I go about proving such a statement? Any help would be great thanks.

- Oct 11th 2009, 02:03 PMPlato
First prove this theorem:

*Every sequence contains a monotone subsequence.*

Then prove that*Every bounded non-increasing (non-decreasing) sequence converges to its least upper bound (greatest lower bound).*

Suppose that $\displaystyle \left( {x_n } \right)$ is a sequence.

Define a set $\displaystyle S = \left\{ {n:\left( {\forall j > n} \right)\left[ {x_j > x_n } \right]} \right\}$.

Take notice of this critical fact: $\displaystyle t \notin S\text{ if and only if }\left( {\exists K > t} \right)\left[ {x_K \leqslant x_t } \right]$.

There are two cases to consider: $\displaystyle S$ is infinite.

In this case there is an increasing subsequence.

If $\displaystyle S$ is finite then there is a non-increasing subsequence. - Oct 11th 2009, 11:48 PMdannyboycurtis
Im still not quite sure how this proves the two statements are equivalent.

Is this proof basically showing that an arbitrary monotone convergent sequence can be shown to be a subsequence of some other bounded sequence? - Oct 12th 2009, 06:19 AMHallsofIvy
Well, that would be trivial! Just add some numbers to your monotone sequence! For example, the monotone sequence $\displaystyle \{x_1, x_2, x_3, ...\}$ is a subsequence of sequence $\displaystyle \{x_1, 1, x_2, 1, x_3, 1, ...\}$

But what you want is the other way- that every sequence contains a monontone subsequence. I'll give you a start: Let S be the set of all indices i such that if j> i, then $\displaystyle a_i< a_j$. In other words, i is in S if and only if $\displaystyle a_i$ is less than all**succeeding**numbers in the sequence. Now, consider two cases:

1) S is infinite.

2) S is either finite or empty. - Oct 12th 2009, 08:43 PMdannyboycurtis
How could I use the Bolzano Wierstrass theorem for sets to prove this?