[proofplan] The proof is an unpacking of the two definitions. Convergence in the uniform metric means that the supremum of the pointwise distances $e(f_k(s),f(s))$ tends to $0$. [Uniform convergence](/page/Uniform%20Convergence) means that these same pointwise distances are eventually small by one index $N$ that works simultaneously for every $s\in S$. Each implication follows by translating one formulation into the other. [/proofplan]

[step:Define the uniform distance controlling all pointwise distances] For each $k\in\mathbb N$, define the non-negative number $a_k$ by \begin{align*} a_k := d_\infty(f_k,f)=\sup_{s\in S} e(f_k(s),f(s)). \end{align*} Because $f_k,f\in B(S,Y)$ and $d_\infty$ is the uniform metric on $B(S,Y)$, each $a_k$ is finite. By the defining property of the supremum, for every $k\in\mathbb N$ and every $s\in S$, \begin{align*} e(f_k(s),f(s))\leq a_k. \end{align*} If $S=\varnothing$, the quantified assertions over $s\in S$ are vacuous and the uniform metric uses the usual convention that the supremum of the empty family of distances is $0$. [/step]

[step:Convert convergence in the uniform metric into uniform convergence]Assume that $f_k\to f$ in the [metric space](/page/Metric%20Space) $(B(S,Y),d_\infty)$. By the definition of convergence in a metric space, for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that, for every $k\in\mathbb N$ with $k\geq N$, \begin{align*} d_\infty(f_k,f)<\varepsilon. \end{align*} Fix such an $\varepsilon>0$ and choose such an $N$. If $k\geq N$ and $s\in S$, then the supremum bound from the previous step gives \begin{align*} e(f_k(s),f(s))\leq d_\infty(f_k,f)<\varepsilon. \end{align*} Thus, for every $\varepsilon>0$, there exists $N\in\mathbb N$ such that $e(f_k(s),f(s))<\varepsilon$ for all $k\geq N$ and all $s\in S$. This is precisely uniform convergence of $f_k$ to $f$ on $S$.[/step]

[guided]We assume that the sequence converges in the metric space $(B(S,Y),d_\infty)$. The definition of metric convergence says that the distance from $f_k$ to $f$ must tend to $0$ in the [real numbers](/page/Real%20Numbers). In this particular metric, that distance is not measured at one point of $S$; it is the largest pointwise error, in the supremum sense: \begin{align*} d_\infty(f_k,f)=\sup_{s\in S} e(f_k(s),f(s)). \end{align*} Let $\varepsilon>0$ be given. Since $f_k\to f$ in the metric $d_\infty$, there exists $N\in\mathbb N$ such that, for every $k\in\mathbb N$ with $k\geq N$, \begin{align*} d_\infty(f_k,f)<\varepsilon. \end{align*} Now fix any $k\geq N$ and any $s\in S$. The number $e(f_k(s),f(s))$ is one of the pointwise distances whose supremum defines $d_\infty(f_k,f)$. Therefore the defining upper-bound property of the supremum gives \begin{align*} e(f_k(s),f(s))\leq d_\infty(f_k,f). \end{align*} Combining this with the choice of $N$ gives \begin{align*} e(f_k(s),f(s))<\varepsilon. \end{align*} The same index $N$ works for every $s\in S$, because the metric $d_\infty$ already took the supremum over all points of $S$. Hence, for every $\varepsilon>0$, there exists $N\in\mathbb N$ such that $e(f_k(s),f(s))<\varepsilon$ whenever $k\geq N$ and $s\in S$. This is exactly uniform convergence on $S$.[/guided]

[step:Convert uniform convergence into convergence in the uniform metric] Assume that $f_k$ converges uniformly to $f$ on $S$. Let $\varepsilon>0$ be given. By uniform convergence applied with $\varepsilon/2>0$, there exists $N\in\mathbb N$ such that, for every $k\in\mathbb N$ with $k\geq N$ and every $s\in S$, \begin{align*} e(f_k(s),f(s))<\varepsilon/2. \end{align*} For each $k\geq N$, the number $\varepsilon/2$ is an upper bound for the set $\{e(f_k(s),f(s)):s\in S\}$ up to strict pointwise inequalities, so its supremum satisfies \begin{align*} d_\infty(f_k,f)=\sup_{s\in S} e(f_k(s),f(s))\leq \varepsilon/2<\varepsilon. \end{align*} Therefore, for every $\varepsilon>0$, there exists $N\in\mathbb N$ such that $d_\infty(f_k,f)<\varepsilon$ for all $k\geq N$. By the definition of convergence in the metric space $(B(S,Y),d_\infty)$, this proves $f_k\to f$ in the uniform metric. [/step]

[step:Conclude the equivalence] The first implication shows that convergence in $(B(S,Y),d_\infty)$ implies uniform convergence on $S$. The second implication shows that uniform convergence on $S$ implies convergence in $(B(S,Y),d_\infty)$. Hence $f_k\to f$ in the uniform metric if and only if $f_k$ converges uniformly to $f$ on $S$. [/step]