[proofplan] The [discrete metric](/page/Discrete%20Metric) turns every nonzero distance in $X$ into the single value $1$. Thus a Lipschitz bound for $f$ is exactly a uniform bound on all distances $d_Y(f(x),f(x'))$ between image points. One direction uses the Lipschitz constant as a diameter bound; the other direction uses the finite image diameter as a Lipschitz constant, with a harmless positive replacement if the diameter is $0$. [/proofplan]

[step:Use a Lipschitz constant to bound the image diameter] Assume that $f$ is Lipschitz. Then there exists a constant $L>0$ such that for every $x,x'\in X$, \begin{align*} d_Y(f(x),f(x'))\leq L\,d_X(x,x'). \end{align*} Let $y,y'\in f(X)$. By definition of the image, there exist $x,x'\in X$ such that $y=f(x)$ and $y'=f(x')$. If $x=x'$, then $d_Y(y,y')=0$. If $x\neq x'$, then the definition of the discrete metric gives $d_X(x,x')=1$, and the Lipschitz inequality gives \begin{align*} d_Y(y,y')=d_Y(f(x),f(x'))\leq L. \end{align*} Therefore every pair of points in $f(X)$ has $d_Y$-distance at most $L$. By the definition of diameter, \begin{align*} \operatorname{diam}_Y(f(X))\leq L<\infty. \end{align*} If $X=\varnothing$, then $f(X)=\varnothing$ and $\operatorname{diam}_Y(f(X))=0$ by the stated convention, so the same conclusion holds. [/step]

[step:Use finite image diameter to construct a Lipschitz constant]Assume that $f(X)$ has finite diameter in $Y$. Define \begin{align*} M:=\operatorname{diam}_Y(f(X)). \end{align*} Then $M<\infty$ and, by the definition of diameter, for every $x,x'\in X$, \begin{align*} d_Y(f(x),f(x'))\leq M. \end{align*} Define the constant $C:=\max\{M,1\}$. Then $C>0$ and $M\leq C$. We prove that $C$ is a Lipschitz constant for $f$. Let $x,x'\in X$. If $x=x'$, then $f(x)=f(x')$, and hence \begin{align*} d_Y(f(x),f(x'))=0=C\,d_X(x,x'). \end{align*} If $x\neq x'$, then $d_X(x,x')=1$, so \begin{align*} d_Y(f(x),f(x'))\leq M\leq C=C\,d_X(x,x'). \end{align*} Thus for every $x,x'\in X$, \begin{align*} d_Y(f(x),f(x'))\leq C\,d_X(x,x'). \end{align*} Therefore $f$ is Lipschitz.[/step]

[guided]Assume that the image $f(X)$ has finite diameter in $Y$. We name this finite bound: \begin{align*} M:=\operatorname{diam}_Y(f(X)). \end{align*} The definition of diameter says that $M$ is a uniform upper bound for the distances between all pairs of points in $f(X)$. Since $f(x)$ and $f(x')$ belong to $f(X)$ for every $x,x'\in X$, we have \begin{align*} d_Y(f(x),f(x'))\leq M \end{align*} for every $x,x'\in X$. To prove that $f$ is Lipschitz, we need a constant multiplying $d_X(x,x')$. Some definitions allow a Lipschitz constant to be $0$, while others require it to be positive, so we choose a positive constant that always works: \begin{align*} C:=\max\{M,1\}. \end{align*} Then $C>0$ and $M\leq C$. Now let $x,x'\in X$. There are two cases, and this is exactly where the discrete metric is used. If $x=x'$, then $f(x)=f(x')$, so \begin{align*} d_Y(f(x),f(x'))=0. \end{align*} Also $d_X(x,x')=0$, hence \begin{align*} d_Y(f(x),f(x'))=0=C\,d_X(x,x'). \end{align*} If $x\neq x'$, then the discrete metric gives $d_X(x,x')=1$. The finite diameter bound gives \begin{align*} d_Y(f(x),f(x'))\leq M. \end{align*} Since $M\leq C$, we obtain \begin{align*} d_Y(f(x),f(x'))\leq C=C\,d_X(x,x'). \end{align*} In both cases, \begin{align*} d_Y(f(x),f(x'))\leq C\,d_X(x,x'). \end{align*} This is precisely the Lipschitz condition for the map $f:X\to Y$, so $f$ is Lipschitz.[/guided]

[step:Conclude the equivalence] The first step proves that every Lipschitz map $f:X\to Y$ has image of finite diameter. The second step proves that finite diameter of $f(X)$ supplies a Lipschitz constant for $f$. Hence $f$ is Lipschitz if and only if $f(X)$ has finite diameter in $Y$. [/step]