Let X dX and Y dY be metric spaces and fXY be an isometry be

Let (X, d_X) and (Y, d_Y) be metric spaces and f:X->Y be an isometry between them. Show that f is a homeomorphism.

Solution

We denote the metric on X as d and the metric on Y by D, to avoid confusion

Proof :

(i) f: X->Y is one-one.

     Let u,v be in X . If f(u) =f(v) , then d(u,v)=D(f(u),f(v)) (as f is an isometry, it preserves distances)

                                                                    =0

implies d(u,v)=0 implying u=v (As d is a metric on X)

Note: We can only assert that f is a homeoemorphism between X and f(X) (the image of X under f). It is not given that f is onto. (for example consider the identity map between Q and R, it is an isometry into R , clearly it is not a homeomorphism)

In what follows , assume that f is onto, so that f^-1 :Y->X is well defined.

f is continuous as the inverse image of fof (any open ball B in Y) = open ball in X of the same radius, because of the isometric property of f.

Reversing the argument , the same is true of the inverse map f^-1 :Y->X.

Thus f is a homeomorphism of the spaces X and Y