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.

  • $\begingroup$ The inverse of an isometry is also an isometry. $\endgroup$ May 1, 2020 at 23:12

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – M A
    May 1, 2020 at 23:26
  • $\begingroup$ @MA6 The inverse is continuous, that's all you need for a homeomorphism. So I don't see your doubt. $\endgroup$ May 1, 2020 at 23:28
  • $\begingroup$ The concept of open sets exists in any metric space and every metric space is also a topological space. A homeomorphism is bijection such that the function and its inverse are both continuous. $\endgroup$ May 1, 2020 at 23:28
  • $\begingroup$ I understand now, thank you! $\endgroup$
    – M A
    May 1, 2020 at 23:34

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