Any bounded seq. has a converging subseq. <=> every bounded monotone seq. converges

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October 11th 2009, 01:14 PM
dannyboycurtis

Any 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.
October 11th 2009, 02:03 PM
Plato

Quote:

Originally Posted by dannyboycurtis

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

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 $\left( {x_n } \right)$ is a sequence.
Define a set $S = \left\{ {n:\left( {\forall j > n} \right)\left[ {x_j > x_n } \right]} \right\}$ .
Take notice of this critical fact: $t \notin S\text{ if and only if }\left( {\exists K > t} \right)\left[ {x_K \leqslant x_t } \right]$ .

There are two cases to consider: $S$ is infinite.
In this case there is an increasing subsequence.

If $S$ is finite then there is a non-increasing subsequence.
October 11th 2009, 11:48 PM
dannyboycurtis

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?
October 12th 2009, 06:19 AM
HallsofIvy

Quote:

Originally Posted by dannyboycurtis

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?

Well, that would be trivial! Just add some numbers to your monotone sequence! For example, the monotone sequence $\{x_1, x_2, x_3, ...\}$ is a subsequence of sequence $\{x_1, 1, x_2, 1, x_3, 1, ...\}$

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 $a_i< a_j$ . In other words, i is in S if and only if $a_i$ is less than all succeeding numbers in the sequence. Now, consider two cases:
1) S is infinite.
2) S is either finite or empty.
October 12th 2009, 08:43 PM
dannyboycurtis

How could I use the Bolzano Wierstrass theorem for sets to prove this?