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

Throughout, $B_{d_X}(x,r)$ and $B_{d_Y}(w,r)$ denote open balls, and we use that a subset is open in a metric space exactly when it contains an open ball about each of its points.

Continuity implies that preimages of open sets are open. Assume $f$ is continuous on $X$, let $V\in\mathcal{T}_{d_Y}$, and let $x\in f^{-1}(V)$. Then $f(x)\in V$, and since $V$ is open in $(Y,d_Y)$ there is a real number $\varepsilon>0$ with $B_{d_Y}(f(x),\varepsilon)\subseteq V$. By continuity of $f$ at $x$ relative to $X$ there is a real number $\delta>0$ such that every $y\in X$ with $d_X(x,y)<\delta$ satisfies $d_Y(f(y),f(x))<\varepsilon$. Let $y\in B_{d_X}(x,\delta)$, so that $d_X(x,y)<\delta$. Then $d_Y(f(y),f(x))<\varepsilon$, and $d_Y(f(x),f(y))=d_Y(f(y),f(x))$ by condition 3 of the definition of a metric, so $f(y)\in B_{d_Y}(f(x),\varepsilon)\subseteq V$ and hence $y\in f^{-1}(V)$. Thus $B_{d_X}(x,\delta)\subseteq f^{-1}(V)$. As $x$ was an arbitrary point of $f^{-1}(V)$, the set $f^{-1}(V)$ is open in $(X,d_X)$, that is, $f^{-1}(V)\in\mathcal{T}_{d_X}$.

Openness of preimages implies continuity. Assume $f^{-1}(V)\in\mathcal{T}_{d_X}$ for every $V\in\mathcal{T}_{d_Y}$. Let $x\in X$ and let $\varepsilon$ be a real number with $0<\varepsilon$. The set $V=B_{d_Y}(f(x),\varepsilon)$ is open in $(Y,d_Y)$ by Open Ball in a Metric Space is Open, so $V\in\mathcal{T}_{d_Y}$ and therefore $U=f^{-1}(V)\in\mathcal{T}_{d_X}$. By condition 2 of the definition of a metric, $d_Y(f(x),f(x))=0$, and $0<\varepsilon$, so $f(x)\in V$ and hence $x\in U$. Since $U$ is open in $(X,d_X)$ there is a real number $\delta>0$ with $B_{d_X}(x,\delta)\subseteq U$. Now let $y\in X$ satisfy $d_X(x,y)<\delta$. Then $y\in B_{d_X}(x,\delta)\subseteq U$, so $f(y)\in V$, that is, $d_Y(f(x),f(y))<\varepsilon$; by condition 3 of the definition of a metric, $d_Y(f(y),f(x))<\varepsilon$. Hence $f$ is continuous at $x$ relative to $X$, and as $x\in X$ was arbitrary, $f$ is continuous on $X$.