Question

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...).

Answer 1

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)<e} is a basis of Y. The function f has been shown to be a bijection and therefore has an inverse f^-1. We also know by the definition of f that d(f^-1(b),f^-1(y)) is equal to d1(b,y). Therefore the inverse image of any arbitrary open ball Bb(e) in Y is an open ball in X which is part of the topological basis induced by the metric d on X.

3) Follows directly from the observations made in 2) about bases and the fact that f preserves the metric: open balls in X are mapped to open balls in Y.

(Obviously f is an isometry...)

[]