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Consider the following definition for a homeomorphism $h$ between metric spaces $(X,d_1)$ and $(Y,d_2)$, with metric topologies $\tau_1$ and $\tau_2$ respectively:

  • Definition: $h: X \rightarrow Y$ is a homeomorphism if $h$ is a bijection between $X$ and $Y$, and the induced map $h_*$ from $\tau_1$ to $\tau_2$ (which is just $h$ redefined to take open sets as input) is a bijection as well. This definition can be found, for example, in this post.

I would like to show that this definition implies "standard continuity" for $h$; that is, that given any $\epsilon > 0$, there exists a $\delta > 0$ such that $d_1(x_1,x_2) < \delta$ implies $d_2(h(x_1),h(x_2)) < \epsilon$ for any such elements $x_1,x_2 \in X$.

Ideally, I would like the argument to be as direct as possible, i.e. avoiding statements that prove $h$ is continuous according to an alternate equivalent definition, which then implies this.

aghostinthefigures
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1 Answers1

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The argument is quite short: Fix some $x \in X$ and let $y=h(x)$. Consider the open ball $B_\epsilon(y)$ around $y \in Y$ for some $\epsilon>0$. Since $h_*$ is a bijection, $h_*^{-1}$ is well-defined from $\tau_2$ to $\tau_1$ and so there exists an open $U \in \tau_1$ with $x\in U$ such that $h_*^{-1}(B_{\epsilon}(y))=U$. Since $U$ is open, it contains an open $B_{\delta}(x)\subset U$ for some $\delta>0$. Clearly $h_*(B_{\delta}(x))\subset B_\epsilon(y)$, which is essentially the definition of continuity you stated. The proof of continuity of the inverse is pretty much identical.

almosteverywhere
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