• anonymous
Someone interested in checking another topology proof for errors/omissions? The proposition is: Let (X,d) and (Y,d1) be two metric spaces. Prove that a function f:(X,d)->(Y,d1) that is surjective and preserves the metric: d(x1,x2) = d1(f(x1),f(x2)) is a homeomorphism. My tentative to prove this is in the comments (as one can only write so much in a question...).
  • Stacey Warren - Expert
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  • schrodinger
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
We have to prove three additional properties of f for it to be a homeomorphism: 1) It must be injective. 2) Continuity of the inverse: The inverse image of f of an open set in Y must be open in X. 3) Continuity of f: The image of an open set in X must be open in Y. Here we go: 1) If f is an injection then for every x1, x2 in X, f(x_1)!=f(x_2). Assume for a contradiction that x1!=x2 and f(x1)=f(x2). Then by the definition of a metric d(x1,x2)>0 and d(f(x1),f(x2))=0. This contradicts our assumption that f preserves the metric. So f must be an injection. 2) We have to show that the inverse image of every open set in Y is open in X. Equivalently we can show that the inverse image of every member of some basis for the topology on Y is a member of a basis for X. Now the set of open balls Bb(e)={y|d1(b,y)

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