[proofplan] The pointwise Cauchy hypothesis gives, for each fixed point $x \in E$, an index after which the values $f_n(x)$ are mutually $\varepsilon$-close. Since $E$ is finite, these finitely many pointwise indices can be replaced by their maximum. That maximum then works simultaneously for every point of $E$, which is exactly uniform Cauchyness. [/proofplan] [step:Handle the empty finite set separately] If $E=\varnothing$, then for any $\varepsilon>0$ we may choose $N=1$. For all $m,n \ge N$ and all $x \in E$, the inequality \begin{align*} d_Y(f_m(x),f_n(x))<\varepsilon \end{align*} holds vacuously because there is no $x \in E$. Hence $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$ in this case. [/step] [step:Choose pointwise Cauchy indices at each point of the finite set] Assume now that $E \ne \varnothing$. Fix $\varepsilon>0$. For each $x \in E$, the sequence $(f_n(x))_{n=1}^{\infty}$ is Cauchy in $(Y,d_Y)$ by hypothesis. Therefore, for each $x \in E$, there exists an index $N_x \in \mathbb{N}$ such that for all $m,n \ge N_x$, \begin{align*} d_Y(f_m(x),f_n(x))<\varepsilon. \end{align*} [guided] Assume $E \ne \varnothing$, and fix an arbitrary $\varepsilon>0$. Our goal is to find one index that works for every point $x \in E$ at once. The hypothesis gives something slightly weaker: for each fixed point $x \in E$, the sequence of values \begin{align*} (f_n(x))_{n=1}^{\infty} \end{align*} is Cauchy in the [metric space](/page/Metric%20Space) $(Y,d_Y)$. Unpacking the definition of a [Cauchy sequence](/page/Cauchy%20Sequence) in a metric space, this means the following. For each fixed $x \in E$ and for the chosen $\varepsilon>0$, there exists an index $N_x \in \mathbb{N}$ such that whenever $m,n \ge N_x$, the two values $f_m(x)$ and $f_n(x)$ are within $\varepsilon$ in the metric $d_Y$: \begin{align*} d_Y(f_m(x),f_n(x))<\varepsilon. \end{align*} The subscript in $N_x$ records that this index may depend on the point $x$. At this stage we have pointwise control, not uniform control, because different points of $E$ may have different indices. [/guided] [/step] [step:Take the maximum of the finitely many pointwise indices] Since $E$ is finite and nonempty, the finite set of natural numbers \begin{align*} \{N_x:x \in E\} \end{align*} has a maximum. Define \begin{align*} N:=\max_{x \in E} N_x. \end{align*} Then $N \in \mathbb{N}$, and for every $x \in E$ we have $N_x \le N$. [/step] [step:Use the maximum index to obtain uniform Cauchy control] Let $m,n \ge N$ and let $x \in E$. Since $N_x \le N$, we have $m,n \ge N_x$. By the defining property of $N_x$, \begin{align*} d_Y(f_m(x),f_n(x))<\varepsilon. \end{align*} Thus for the fixed $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $m,n \ge N$ and all $x \in E$, \begin{align*} d_Y(f_m(x),f_n(x))<\varepsilon. \end{align*} Because $\varepsilon>0$ was arbitrary, $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$. [/step]