= Solution
The <continuity> condition at $x\in X$ between <metric spaces> is
$$
\boxed{\forall\epsilon>0\ \exists\delta>0:\quad d(x,y)<\delta\Longrightarrow e(f(x),f(y))<\epsilon.}
$$
The function is continuous when this holds at every $x$.
If $f$ is continuous and $U\subseteq Y$ is an <open set>, take $x\in f^{-1}(U)$. Some $\epsilon$-ball around $f(x)$ lies in $U$. Continuity gives a $\delta$-ball around $x$ mapped into that ball and hence contained in $f^{-1}(U)$. Thus $f^{-1}(U)$ is open, including the empty set, in the notation for the <image and preimage of a function>.
Conversely, suppose every open set has open preimage. For fixed $x$ and $\epsilon>0$, the preimage of the open ball $B_e(f(x),\epsilon)$ is open and contains $x$, so contains some $B_d(x,\delta)$. This gives the displayed metric definition. Therefore $\boxed{f\text{ is continuous}\iff f^{-1}(U)\text{ is open for every open }U}$.
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