This is a nice problem, compared to the previous one, at least it seems so.
One needs to show that a metric space (X, d) is complete iff for every nested sequence ... [tex]\subseteq[/tex]A2[tex]\subseteq[/tex]A1 of nonempty closed subsets of X such that diam An --> 0, the intersection of the sets An is non-empty.
The Attempt at a Solution
I'll start with ==>.
The basic idea is to construct a Cauchy sequence in X such that its limit lies in all of the sets Ai.
Define the sequence xn as follows:
For ε = 1, choose x1 from the set An1, where n1 is the integer such that n >= n1 implies diam An < 1. For ε = 1/2, choose x2 from An2 in the same manner, and for ε = 1/n in general, choose xn.
I claim xn is a Cauchy sequence. Let ε > 0 be given. Choose N so that 1/N < ε. Clearly, for all N1, N2 >= N, we have d(xN1, xN2) < 1/n < ε.
Since X is complete, xn converges to a limit x. I assert that x lies in [tex]\cap[/tex]An.
Now, this is a part I'm not 100% sure about.
First, as xn --> x in A1, and A1 is a closed subset of a complete space, which makes A1 complete, the limit x lies in A1.
Then, if we look at all the members of the sequence xn except for those which lie in A1 (there must be only finitely many such members!), we have a subsequence xn' of our original sequence xn. Now, this sequence converges to the same limit x as xn, by Lemma 43.1. in Munkres, since if xn' --> y would hold, then xn --> y, which contradicts the fact that the limit is unique. Hence, x lies in A2. Inductively, x must lie in An, for every n.
Does this work?