Conservation of completeness by uniformly continuous bijection

[h-simplex]

Homework Statement

I want to prove this proposition:

Let [itex] f: M \rightarrow N [/itex] be a uniformly continuous bijection between metric spaces. If M is complete, then N is complete.

The Attempt at a Solution

I have a 'partial' solution, whose legitimacy hinges upon a claim that I am unable to prove, namely that the inverse of a uniformly continuous bijection is uniformly continuous.

Assuming that is true, here goes my 'proof':
Let [itex] (y_n)[/itex] be a Cauchy sequence in N. Then, since [itex]f^{-1}[/itex] is uniformly continuous, [itex](f^{-1}(y_n))[/itex] is Cauchy in M; since M is complete, there exists [itex]a \in M[/itex] such that [itex]f^{-1}(y_n) \rightarrow a[/itex]. Hence by continuity of f, we have [itex]f(f^{-1}(y_n)) = y_n \rightarrow f(a) \in N[/itex]. Thus N is complete.

The above proof seems very 'natural' ... but as I mentioned, it hinges on something I am unable to prove, or disprove. So is my claim correct and if so how can I prove it? If not... well then I have no idea how to prove the original claim; this was my best idea.

 

[micromass]

I would try to find a counterexample for both statements (that is: "uniform continuous images of complete spaces are complete" and "uniform continuous bijections are uniform isomorphisms"). I doubt that they are true.