Let be a sequence such that for all then since for all and are closed and bounded there is a subsequence with such that . And by construction, and since the sequence is nested, for all .
Suppose that for each we have a non-empty closed and bounded set and
Prove that is non empty. [Hint: use Bolzano-Weierstrass]
Solution:
Let . Then there exists a bounded sequence with a convergent subsequence whose limit is . I need to show that . I can see that since . But how do I show that if then ? Any help would be appreciated.
I have thought about this, does the way I have written this make sense and is it a valid proof?
Let be a sequence such that (i.e. ... )
such that converges to limit since closed.
Since the sets are nested the sequence is in with subsequence which converges to x as is a subsequence of a convergent sequence (the first m terms have simply gone).
Hence and so .