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*.

Answer 1

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?

Answer 2

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*.

Answer 3

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.

Answer 4

* \( (s_n) \)

Answer 5

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