Proof of Continuity of a Map Between Metric Spaces via Preimages of Open Sets

Throughout, membership in an open ball is as in Open Ball in a Metric Space §membership, preimages are as in Sets and Maps: Ordinary Notation §images, and $d_{X}$ and $d_{Y}$ are symmetric by Metric Space §symmetric.

Continuity implies that preimages of open sets are open. Assume $f$ is continuous, let $V\subseteq Y$ be open in $(Y,d_{Y})$, and let $x\in f^{-1}(V)$, so that $f(x)\in V$. By Open Subset of a Metric Space §open there is a real $\varepsilon>0$ with $B_{d_{Y}}(f(x),\varepsilon)\subseteq V$. By Continuous Map Between Metric Spaces §continuous, $f$ is continuous at $x$, so, $\varepsilon$ being chosen, Continuous Map Between Metric Spaces §continuous-at gives a real $\delta>0$ such that every $y\in X$ with $d_{X}(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$. Let $y\in B_{d_{X}}(x,\delta)$; then $y\in X$ and $d_{X}(x,y)<\delta$, so $d_{Y}(f(x),f(y))=d_{Y}(f(y),f(x))<\varepsilon$, whence $f(y)\in B_{d_{Y}}(f(x),\varepsilon)\subseteq V$ and $y\in f^{-1}(V)$. Thus $B_{d_{X}}(x,\delta)\subseteq f^{-1}(V)$. As $x$ was an arbitrary point of the subset $f^{-1}(V)$ of $X$, $f^{-1}(V)$ is open in $(X,d_{X})$ by Open Subset of a Metric Space §open.

Openness of preimages implies continuity. Assume that $f^{-1}(V)$ is open in $(X,d_{X})$ for every subset $V$ of $Y$ open in $(Y,d_{Y})$. Let $x\in X$, let $\varepsilon>0$ be real, and put $V=B_{d_{Y}}(f(x),\varepsilon)$.

By Open Ball in a Metric Space is Open §open, $V$ is open in $(Y,d_{Y})$.

Hence $U=f^{-1}(V)$ is open in $(X,d_{X})$ by assumption. Since $d_{Y}(f(x),f(x))=0<\varepsilon$ by Metric Space §zero, $f(x)\in V$ and $x\in U$. By Open Subset of a Metric Space §open there is a real $\delta>0$ with $B_{d_{X}}(x,\delta)\subseteq U$. Let $y\in X$ with $d_{X}(x,y)<\delta$. Then $y\in B_{d_{X}}(x,\delta)\subseteq U$, so $f(y)\in V$, that is, $d_{Y}(f(x),f(y))<\varepsilon$, and $d_{Y}(f(y),f(x))<\varepsilon$ by symmetry. As $\varepsilon$ was arbitrary and $\delta$ was chosen after it, $f$ is continuous at $x$; as $x\in X$ was arbitrary, $f$ is continuous by Continuous Map Between Metric Spaces §continuous.