Let be an open set. Suppose that the map is a and is .
Prove that .
Not true! For example, there exists a bijection between and (and in fact a homeomorphism). Infinite subsets of infinite sets can often be put into bijective correspondance with their supersets.
The crucial piece of information here is that the function is in fact uniformly continuous. I'm still working on figuring out what that entails.
I think I have it. Suppose . Then is not closed (since only closed and open sets are the empty and whole sets). In particular there exists a such that there exists a sequence with each . Then since the sequence is convergent, it is Cauchy, and since the function is uniformly continuous the sequence of images is Cauchy. Since is complete, there exists such that . Since the function was onto, there exists u such that .
Then since is continuous, implies . But so by uniqueness of limits . Since we assumed but , we have our contradiction!
Yes! The proof is easy too. Suppose . Let be given. Then by the definition of convergence there exists a such that for all , . Then for all , .
Convergence is a stronger condition than Cauchy since to be convergent we must identify the element the sequence converges to, whereas with cauchy we only know the sequence elements get closer together.