[proofplan] We prove the equivalence directly from the $\varepsilon$-$\delta$ definition of continuity for metric spaces. If $f$ is continuous, then near any point mapping to $y$, the choice \begin{align*} \varepsilon &= \frac{1}{2} \end{align*} forces nearby points to map exactly to $y$, so the fiber over $y$ is open. Conversely, if every fiber is open, then for each point $x_0\in X$ the open fiber through $x_0$ supplies a radius on which $f$ is constant; this gives the $\varepsilon$-$\delta$ condition for the [discrete metric](/page/Discrete%20Metric). [/proofplan] [step:Use continuity to show that each point fiber is open] Assume that $f:(X,d_X)\to (Y,d_Y)$ is continuous. Fix $y\in Y$ and define the fiber \begin{align*} A_y := f^{-1}(\{y\}) = \{x\in X : f(x)=y\}. \end{align*} We prove that $A_y$ is open in $X$. Let $x_0\in A_y$. Since $f$ is continuous at $x_0$, applying the $\varepsilon$-$\delta$ definition with \begin{align*} \varepsilon &= \frac{1}{2} \end{align*} gives a number $\delta_{x_0}>0$ such that, for every $x\in X$, \begin{align*} d_X(x,x_0)<\delta_{x_0} \implies d_Y(f(x),f(x_0))<\frac{1}{2}. \end{align*} Because $x_0\in A_y$, we have $f(x_0)=y$. In the discrete metric, the only distances strictly smaller than $\frac{1}{2}$ are zero, so $d_Y(f(x),f(x_0))<\frac{1}{2}$ implies $f(x)=f(x_0)=y$. Hence \begin{align*} B_X(x_0,\delta_{x_0}) \subset A_y, \end{align*} where $B_X(x_0,\delta_{x_0}) := \{x\in X : d_X(x,x_0)<\delta_{x_0}\}$ is the open ball in $(X,d_X)$. Thus every point of $A_y$ is contained in an open ball lying inside $A_y$, so $A_y$ is open in $X$. Since $y\in Y$ was arbitrary, $f^{-1}(\{y\})$ is open in $X$ for every $y\in Y$. [/step] [step:Use open fibers to prove the epsilon-delta condition] Assume conversely that $f^{-1}(\{y\})$ is open in $X$ for every $y\in Y$. We prove that $f$ is continuous at every point of $X$. Fix $x_0\in X$ and define \begin{align*} y_0 := f(x_0). \end{align*} The fiber \begin{align*} A_{y_0} := f^{-1}(\{y_0\}) \end{align*} is open in $X$ by hypothesis, and $x_0\in A_{y_0}$ by definition of $y_0$. Therefore there exists a radius $r_{x_0}>0$ such that \begin{align*} B_X(x_0,r_{x_0}) \subset A_{y_0}. \end{align*} Let $\varepsilon>0$ be arbitrary. Define \begin{align*} \delta := r_{x_0}. \end{align*} If $x\in X$ satisfies $d_X(x,x_0)<\delta$, then $x\in A_{y_0}$, so $f(x)=y_0=f(x_0)$. Hence \begin{align*} d_Y(f(x),f(x_0))=0<\varepsilon. \end{align*} This proves continuity of $f$ at $x_0$. Since $x_0\in X$ was arbitrary, $f$ is continuous on $X$. [guided] We assume that every point fiber is open and want to prove metric continuity. The relevant continuity condition is local at a point: for each $x_0\in X$ and each $\varepsilon>0$, we must find a radius $\delta>0$ such that points within $\delta$ of $x_0$ have images within $\varepsilon$ of $f(x_0)$. Fix $x_0\in X$. The value of $f$ at this point is an element of $Y$, so define \begin{align*} y_0 := f(x_0). \end{align*} Now consider the fiber over this value: \begin{align*} A_{y_0} := f^{-1}(\{y_0\}) = \{x\in X : f(x)=y_0\}. \end{align*} By the hypothesis of this direction, $A_{y_0}$ is open in $X$. Also $x_0\in A_{y_0}$, because $f(x_0)=y_0$ by definition. The definition of openness in the [metric space](/page/Metric%20Space) $(X,d_X)$ now gives a radius $r_{x_0}>0$ such that the open ball \begin{align*} B_X(x_0,r_{x_0}) := \{x\in X : d_X(x,x_0)<r_{x_0}\} \end{align*} satisfies \begin{align*} B_X(x_0,r_{x_0}) \subset A_{y_0}. \end{align*} This is the key point: inside this ball, the function $f$ is not merely close to $f(x_0)$; it is equal to $f(x_0)$. Let $\varepsilon>0$ be arbitrary and set \begin{align*} \delta := r_{x_0}. \end{align*} If $x\in X$ satisfies $d_X(x,x_0)<\delta$, then $x\in B_X(x_0,r_{x_0})$, hence $x\in A_{y_0}$. Therefore $f(x)=y_0=f(x_0)$, and the discrete metric gives \begin{align*} d_Y(f(x),f(x_0))=0<\varepsilon. \end{align*} Thus the $\varepsilon$-$\delta$ condition holds at $x_0$. Since the point $x_0\in X$ was arbitrary, $f$ is continuous on $X$. [/guided] [/step] [step:Combine the two implications] The first step proves that continuity of $f$ implies openness of every fiber $f^{-1}(\{y\})$. The second step proves that openness of every fiber implies continuity of $f$. Therefore $f:(X,d_X)\to(Y,d_Y)$ is continuous if and only if $f^{-1}(\{y\})$ is open in $X$ for every $y\in Y$. [/step]