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Reminder: An isometry of metric spaces $(X,d)$, $(Y,d')$ is a function

$$f:X \to Y$$ Such that for any $s,t \in X$ we have

$$d(s,t) = d(f(s),f(t))$$

i.e. It preserves all distances.

A homeomorphism of metric spaces $(X,d)$, $(Y,d')$ is a bijection from $X$ to $Y$ which is continuous and has continuous inverse.

Although every surjective isometry is a homeomorphism, it is not true that every homeomorphism is an isometry, however if we are given a metric space $(X,d)$ and two subpaces $(S,d)$ and $(K,d)$ of $X$, and we know that for any other subset $J \subseteq X$, $J\cap S$ is homeomorphic to $J\cap K$, then there is an isomemtry $S\to K$ (In fact one can simply conclude that $S = K$) However, we can try to strengthen this theorem and assume that for every open set $J\subseteq X$, there is a homeomorphism $J\cap S \to J\cap K$.

Question: Is this still sufficient to conclude that there is an isometry between $S$ and $J$ ?

Carlyle
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No. An example is $X={\mathbb R}$ with the standard metric, while $S={\mathbb Q}$, $K=S\setminus \{0\}$.

Moishe Kohan
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  • I can see that this might be a counterexample, but it isn't immediately clear to me how one would define the homeomorphism from say $(-1,1)\cap K$ to $(-1,1) \cap S$, is it possible for you to elaborate on this? – Carlyle Dec 15 '23 at 22:28
  • @Carlyle: This was discussed many times on MSE, e.g. https://math.stackexchange.com/questions/68946/is-every-countable-dense-subset-of-mathbb-r-ambiently-homeomorphic-to-mathb – Moishe Kohan Dec 16 '23 at 00:42