## Challenging Problem!!! (Equivalent Metrics)

I need to show that if (X,p) is a non-compact metric space, then there exists a metric p* equivalent to p such that (X,p*) is not complete.

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 Blog Entries: 1 Recognitions: Homework Help It helps us if we know where you're getting stuck... nobody's going to just solve the problem for yo. I'd start by writing down the definition of compactness, and picking an appropriate cover to demonstrate the opposite. See what you can do with that
 I always find it easier to work with a specific example and generalise.. e.g. take the interval (0,1) on the real line with standard metric - what are it's features, why is it not complete? Why is (-infinity,infinity) complete when the two sets are topologicaly the same?

## Challenging Problem!!! (Equivalent Metrics)

OK here is an example. Take the set of the natural numbers. Under the usual metric p it is non-compact. But now I can define a new metric p* by p*(n,n)=0 and p*(n,n+1)=1/2^n. It is easy to check that p and p* are equivalent. Further, the sequence {n} used to be unbounded under p, but it is now a Cauchy sequence that does not converge under p*. Thus, the set of naturals under p* is not complete.

To generalize this, the idea is to assume WLOG that (X,p) is complete but not totally bounded. So we can take an unbounded sequence in (X,p), and make it Cauchy under p* by somehow contracting the distances. Since it was unbounded under p, intuitively it will be non-convergent under p*. I've been trying since a week to find such p*.
 How can you assume wlog of that X is unbounded? What about X = space of continuous functions from R to R which are bounded by 1, and p is uniform metric p(f,g)=sup|f(x)-g(x)| That is a complete and bounded space, but not compact. Can you give an equivalent non-complete metric in this case?

 Quote by gel How can you assume wlog of that X is unbounded? What about X = space of continuous functions from R to R which are bounded by 1, and p is uniform metric p(f,g)=sup|f(x)-g(x)| That is a complete and bounded space, but not compact. Can you give an equivalent non-complete metric in this case?
I assume it is not TOTALLY bounded by Heine-Borel. Not compact=> not complete or not totally bounded. If not complete, then we are done. Therefore assume not totally bounded but complete.
 ok, so it is not totally bounded. Your post confused me by stating that there exists an unbounded sequence. Not being totally bounded means that for some r>0, it cannot be covered by finitely many balls of radius r. So, there is a sequence xn with p(xm,xn)>r for all m != n. However, it does not mean that the sequence is unbounded.
 Given such a sequence, if you can deform p to give a new (and equivalent) metric p* under which xn is Cauchy, then that would be enough.
 Note that every infinite discrete set admits a bounded incomplete metric. Given noncompact (X,p), find e s.t. the set of balls of radius e has no finite subcover. Let {Bi} (indexed by a set I) be a minimal subcover (note we use Zorn's lemma here. This is simply to make the proof cleaner.) Assign the set I a bounded incomplete metric (say q(b1,b2)) for the discrete topology. For any index i, define m(i)=inf(q(i,i')), taken over all i' in I. If x1, x2 are in at least one ball Bi together, define p*(x1, x2)=(1/2e)*p(x1,x2)*(sup(m(i)), the sup being taken over all i s.t. Bi contains both x1 and x2. Otherwise define p*(x1,x2)=sup(q(i1,i2)), taken over all i1 with Bi1 containing x1 and i2 with Bi2 containing x2. I don't have time to check thoroughly right now, but I think this should be a metric of the type you're looking for. Seems like an awful lot of work though, I'd like to see an easier way.
 given any symmetric function f:X x X -> R with f(x,x)=0, you can define a pseudo-metric $$p^*(x,y) = \inf\sum_{k=1}^{n} f(x_{k-1},x_k)$$ where the inf is over all sequences x0,x1,...,xn with x0=x, xn=y. You can define f(x,y) to equal what you want on some chosen sequence and p elsewhere. I think you should be able to force a given non uniformly bounded sequence to be Cauchy, and so incomplete, in this way.
 Thanks, gel. That construction is certainly much easier than mine (though they work the same way).