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If a set $A$ is isometric with a set $B$, does that mean there exists an isometry from $A$ to $B$? Or is it the other way around, that there exists an isometry from $B$ to $A$?

Edit: I forgot that there were multiple definitions of an isometry, but my book's definition does not require it to be surjective. Sorry about that.

2 Answers2

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It means that there is a surjective isometry from $A$ to $B$ and it also means that there as a surjective isometry from $B$ to $A$. These assertions are equivalent.

José Carlos Santos
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The answer is yes to both questions. Any isometry is a bijection onto its range, and the inverse is also a bijection. This means that there exists an isometry such that $f(A)=B$ if and only if there exists an isometry $g$ such that $g(B)=A$.

N. S.
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  • I have updated my question. Sorry about that. –  Sep 12 '17 at 23:52
  • @Jake It makes no difference. If $f : V_1 \to V_2$ is an isometry, then $f: V_1 \to Im(f) \subset V_2$ is a bijective isometry. If $f$ takes $A$ into $B$ then $B=f(A) \subset Im(f)$. – N. S. Sep 12 '17 at 23:55
  • @Jake To spell out the argument, $f^{-1} : Im(f) \to V_1$ is an isometry which takes $B$ to $A$. – N. S. Sep 12 '17 at 23:57
  • @Jake Or if your definition of isemetry is given at the level of sets $A,B$, not as an isometry of larger spaces which takes $A$ into $B$, then isometric sets MUST mean onto isometry between $A$ and $B$. – N. S. Sep 12 '17 at 23:59
  • Mine is specifically with sets that are metric spaces, but I didn't know it made a difference. –  Sep 13 '17 at 00:23