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Conservation of completeness by uniformly continuous bijection

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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.
  2. jcsd
  3. Feb 5, 2013 #2
    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.
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