I cannot figure out metric spaces. Let (S, d) and (S*, d*) be metric spaces. Show that if F:S-S* is uniformly continuous, and if (sn) is a Cauchy sequence in S, then (f(sn)) is a Cauchy sequence in S*.

Question

I cannot figure out metric spaces.  Let (S, d) and (S*, d*) be metric spaces.  Show that if F:S->S* is uniformly continuous, and if (sn) is a Cauchy sequence in S, then (f(sn)) is a Cauchy sequence in S*.

jamesj · Answer

By definition of uniformly continuous for any pair $ x, y \in S* $ and $ \epsilon > 0 $ there exists a $ \delta > 0 $ such that $$ d(x,y) < \delta \ \implies d*(f(x),f(y)) < \epsilon $$ What's very useful about uniform continuity, is that the choice of $ \delta $ doesn't depend on the x and y. It only depends on the epsilon. Now, given that, and given the definition of Cauchy, can you begin to see how to construct a proof?

anonymous · Answer

So, from the problem we're saying that the function f maps from S to S*, and the function f is continuous.  Thus, a Cauchy sequence sn in S is mapped by the function f, f(sn), to S*.

jamesj · Answer

Perhaps a counter example will help you.  Let S and S* be the positive reals with the usual metric.  Define f by f(x) = 1/x.  Let $ s_n) $ be the Cauchy sequence $ s_n = 1/n $.  Then clearly the sequence $ (f(s_n)) = (n) $ is not Cauchy.  But then $ f $ isn't uniformly continuous.

jamesj · Answer

* $ (s_n) $

anonymous · Answer

Thank you for your time. I'm reading through my book and your suggestions above to try to work this out.