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## Nested intervals and the Bolzano-Weierstrass Theorem

Let be a bounded sequence, so for some positive real and for all . Then any subsequence of is bounded by the same number , since . For each , define the subsequence of by the formula:

Note that . Also for each , we have : in fact is the first term of the subsequence . Note that each set is bounded by , so each has a supremum and an infimum. Also from the definition of , we see that , whenever . For each define:

Then by definition of and and by the fact that , we have the inequality, valid for each :

Also, when , since , we have and .
So the sequence is monotonic increasing and the sequence is monotonic decreasing.
Define , for each .
Then we have just shown that is a nested sequence of closed bounded intervals, since the sequence is increasing, the sequence is decreasing and we have , for all . Put .
Then by the theory of nested intervals, we have , where and and , with .
By definition, we write:

The first observation is that if , so the sequences and have the same limit, then by squeeze applied to the inequality , valid for all , we get that the sequence converges with limit .

Next we construct a subsequence of with limit .
Let be a sequence of positive real numbers, with limit zero. For example we could take , for each .
• Put .
First, since is less than the supremum of the set , there is an , so , such that:

• Pick an integer .
Then since is less than the supremum of , there is an , with , such that:

• Pick an integer .
Then since is less than the supremum of , there is an , with , such that:

We continue forever, yielding sequences and such that each sequence is strictly increasing and they are interwoven in the sense that , for each and such that we have, for each , the inequality:

Then the sequence is a subsequence of the sequence , so converges with limit . So, by squeeze applied to the inequality , valid for any , we have that the subsequence also converges, with the same limit . Similarly we construct a subsequence of the sequence with limit .
Let be a sequence of positive real numbers, with limit zero.
• Put .
First, since is greater than the infimum of , there is an , so , such that:

• Pick .
Then since is greater than the infimum of , there is an , with , such that:

• Pick .
Then since is greater than the infimum of , there is an , with , such that:

We continue forever, yielding sequences and such that each sequence is strictly increasing and they are interwoven in the sense that , for each and such that we have, for each , the inequality:

Then the sequence is a subsequence of the sequence , so converges with limit . So, by squeeze applied to the inequality , valid for any , we have that the subsequence also converges, with the same limit .

So we proved that the sequence has two subsequences, one with limit and another with limit . Consequently, if , the sequence cannot be convergent.

So we have proved:
• A bounded sequence is convergent if and only if and then its limit is .
Also we have proved:
• The Bolzano-Weierstrass Theorem: every bounded sequence has a convergent subsequence.
In fact we have just constructed one or two such subsequences, one with limit the other with limit .

Next let be any subsequence of .
Then the sequence is a subsequence of the sequence , since , for some and , for each .
So and , so we have:
and .
In particular, if is convergent, its limit is , so we have proved:
• If is a subsequence of a bounded sequence , then we have the inequality:

• If is a convergent subsequence of a bounded sequence , then its limit satisfies the inequality:

Finally we have the result:
• Let be a bounded sequence. Suppose that every convergent subsequence of has the same limit . Then is also convergent with limit .
This follows since we have shown that there is a convergent subsequence of with limit and another convergent subsequence of with limit so these limits must be equal and equal to , so by the above results, since , we have that converges with limit .

The converse of course is true also: if a sequence converges with limit then all subsequences of converge with the same limit .

Next: Two other proofs of Up: Thetheoryofnestedintervals Previous: Nested intervals
George A. J. Sparling 2012-04-10