# Show that if two metric spaces are isometrically isomorphic then the induced topological spaces are homeomorphic.

Question: Show that if two metric spaces are isometrically isomorphic then the induced topological spaces are homeomorphic.

We need to show that there exists a homeomorphism $$f:X\to Y$$ between the induced topological spaces, i.e. a continuous bijection such that $$f^{-1}$$ is continuous.

My idea was to first prove such a map exists between the open sets of $$X$$ and $$Y$$ because then such a map must exist between the induced topological spaces (since a topological space is just a set along with its 'open' sets), and so we would be done. Is this all correct?

My attempt so far:

Let $$(X,d)$$ and $$(Y,p)$$ be two isometrically isomorphic metric spaces. Then there exists a bijective isometry $$f:X\to Y$$ with $$d(x,y)=p(f(x), f(y))$$. Note that $$f$$ is continuous (simply take $$\delta=\epsilon$$ in the definition of continuity). And now im stuck. I guess from here we need to show the inverse of f is continuous, but im not sure how..

Any help would be appreciated.

• The inverse of an isometry is also an isometry. – rubikscube09 May 1 at 23:12

The inverse is also an isometry: $$d(f^{-1}(u), f^{-1}(v)) =p(u,v)$$ as seen by just putting $$u=f(x)$$ and $$v=f(y)$$. Hence the inverse is also continuous.
• Ahh yes that makes sense. Im also a bit unsure how to phrase the 'conclusion' of this proof.. how can I say that if f has these properties when acting on metric spaces then it must have these when X and Y are topological spaces? Would this suffice: Now I've shown $f:X\to Y$ is bijective, continuous and $f^{-1}:Y\to X$ is also continuous. Hence these properties also hold on the open sets of $X$ and $Y$. Since the induced topological spaces are just the 'open' sets of $(X,d)$ and $(Y,p)$ we have that these properties also hold when X and Y are topological spaces. – MA 6 May 1 at 23:26