Completeness and nested sequences

[radou]

Homework Statement

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 ... $$\subseteq$$A2$$\subseteq$$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 $$\cap$$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?

 

[micromass]

Yeah, that seems alright!

 

[radou]

Yeah, that seems alright!

OK, thanks. The idea for the other direction is that if xn is a Cauchy sequence, we can find a nested sequence of non-empty closed sets of specific diameters containing elements of this sequence, and somehow to show that xn must converge to an element of their non-empty intersection.

I'll think about this tomorrow.

 

[radou]

Here's an attempt for direction "<==".

Let xn be a Cauchy sequence in X. For ε = 1, take N1 such that for all m, n >= N1 we have d(xm, xn) < 1, so all but finitely many elements of the sequence lie in the set Cl(B(xN1, 1)). Now for ε = 1/2, take N2 and in the same manner arrive at a set Cl(B(xN2, 1/2)), which again contains all but finitely many elements of the sequence xn. Proceed inductively. Now, define A1 = Cl(B(xN1, 1)), A2 = Cl(B(xN1, 1))$$\cap$$Cl(B(xN2, 1/2)), and so on (I'll omit the general case because of potentially confusing notation - sorry, but I'm too lazy to TeX). Now, every set from the collection {An} is closed, and we have diam An --> 0. By hypothesis, the intersection of these sets is non-empty.

Now I have a dilemma.

Let L be an element of the intersection. I assert xn --> L. Indeed, let ε > 0 be given. Choose N such that 1/N < ε. There exists a set of the collection {An} with diameter less than 1/N, and hence for all n >= N, d(xn, L) < ε must hold.

The point is - I don't really know L is the only element from the intersection $$\cap$$An. So I can't just "choose" it and say that xn --> L. On the other hand, since X is Hausdorff, xn can converge to at most one limit.

Thanks in advance.

 

[losiu99]

If diam An --> 0, then clearly there is only one point of intersection. Should x1, x2 be in the intersection, their distance would be arbitrarily small, since x1, x2 are elements of An for every n, d(x1, x2) is bounded by diam An. If x1 =\= x2, their distance is positive, and so diam An --\-->0.

 

[radou]

If diam An --> 0, then clearly there is only one point of intersection. Should x1, x2 be in the intersection, their distance would be arbitrarily small, since x1, x2 are elements of An for every n, d(x1, x2) is bounded by diam An. If x1 =\= x2, their distance is positive, and so diam An --\-->0.

Ahh, this is so simple, I should have seen that more clear! Thanks!