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We defined an isometry to be a bijection $f:X\rightarrow X'$ such that $d'(f(x_1),f(x_2))=d(x_1,x_2)$ $\forall x_1,x_2\in X$. Show that any isometry is a homeomorphism.

So my definition of homeomorphism is that a function $f:X\rightarrow X'$ is a homeomorphism if $f$ is a bijection and $f^{-1}$ is continuous. So I have to show that

(a) $f$ is continuous.

$\forall\epsilon>0$ pick $\delta=f^{-1}(\epsilon)$. Then it follows that $d(x_1,x_2)<\delta\implies d'(f(x_1),f(x_2))<\epsilon.$

(b) $f^{-1}$ is continuous. Is this just a reverse of (a)?

  • 2
    $\delta=f^{-1}(\epsilon)$ makes no sense. It seems you are making things way too complicated.2012-03-15
  • 2
    That is not the definition of "homeomorphism". A homeomorphism is a *continuous* bijection whose inverse is continuous.2012-03-15

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