[proofplan] We follow the standard three-step procedure for proving completeness of function spaces. First, the Cauchy condition in the sup-metric implies pointwise Cauchy, and completeness of the target space $(Y, e)$ produces a pointwise limit $f$. Second, we show $f$ is bounded by comparing it to a single bounded $f_N$. Third, we show $\mathcal{D}(f_n, f) \to 0$ by passing to the limit in the uniform Cauchy estimate. [/proofplan] [step:Construct the pointwise limit using completeness of $Y$] Let $(f_n)$ be a Cauchy sequence in $(\ell_\infty(S, Y), \mathcal{D})$. Fix $s \in S$. For all $m, n \in \mathbb{N}$: \begin{align*} e(f_m(s), f_n(s)) \leq \sup_{t \in S} e(f_m(t), f_n(t)) = \mathcal{D}(f_m, f_n). \end{align*} Since $(f_n)$ is Cauchy in the sup-metric, $\mathcal{D}(f_m, f_n) \to 0$ as $m, n \to \infty$, so $(f_n(s))$ is Cauchy in $(Y, e)$. Since $Y$ is complete, $f_n(s)$ converges to some limit. Define $f: S \to Y$ by $f(s) = \lim_{n \to \infty} f_n(s)$. [/step] [step:Verify that the limit function $f$ is bounded] Since $(f_n)$ is Cauchy in the sup-metric, there exists $N \in \mathbb{N}$ with $\mathcal{D}(f_m, f_n) < 1$ for all $m, n \geq N$. Fix $s_0 \in S$. For any $s \in S$ and $n \geq N$: \begin{align*} e(f_n(s), f_N(s)) \leq \mathcal{D}(f_n, f_N) < 1. \end{align*} Passing to the limit $n \to \infty$ (using continuity of $e$ in its first argument: $f_n(s) \to f(s)$ gives $e(f_n(s), f_N(s)) \to e(f(s), f_N(s))$): \begin{align*} e(f(s), f_N(s)) \leq 1. \end{align*} Since $f_N$ is bounded, there exists $M > 0$ with $e(f_N(s), f_N(s_0)) \leq M$ for all $s \in S$. The triangle inequality gives \begin{align*} e(f(s), f_N(s_0)) \leq e(f(s), f_N(s)) + e(f_N(s), f_N(s_0)) \leq 1 + M \end{align*} for all $s \in S$, so $f$ is bounded and hence $f \in \ell_\infty(S, Y)$. [/step] [step:Prove that $f_n \to f$ in the sup-metric] Fix $\varepsilon > 0$. Since $(f_n)$ is Cauchy, there exists $N \in \mathbb{N}$ with $\mathcal{D}(f_m, f_n) < \varepsilon$ for all $m, n \geq N$. Fix $n \geq N$ and $s \in S$. For all $m \geq N$: \begin{align*} e(f_m(s), f_n(s)) \leq \mathcal{D}(f_m, f_n) < \varepsilon. \end{align*} Taking $m \to \infty$ and using continuity of $e$: since $f_m(s) \to f(s)$, we obtain \begin{align*} e(f(s), f_n(s)) \leq \varepsilon. \end{align*} Since this holds for all $s \in S$: \begin{align*} \mathcal{D}(f_n, f) = \sup_{s \in S} e(f_n(s), f(s)) \leq \varepsilon \quad \text{for all } n \geq N. \end{align*} Therefore $\mathcal{D}(f_n, f) \to 0$, and $(\ell_\infty(S, Y), \mathcal{D})$ is complete. [guided] The critical step is passing from the Cauchy estimate $e(f_m(s), f_n(s)) < \varepsilon$ to the limit estimate $e(f(s), f_n(s)) \leq \varepsilon$. This requires continuity of the metric $e$. **Why is $e$ continuous?** For any fixed $z \in Y$, the function $y \mapsto e(y, z)$ is continuous because the reverse triangle inequality gives $|e(y_m, z) - e(y, z)| \leq e(y_m, y)$. If $y_m \to y$, then $e(y_m, y) \to 0$, so $e(y_m, z) \to e(y, z)$. **Passing to the limit:** Fix $n \geq N$ and $s \in S$. For all $m \geq N$, we have $e(f_m(s), f_n(s)) \leq \mathcal{D}(f_m, f_n) < \varepsilon$. As $m \to \infty$, $f_m(s) \to f(s)$ (pointwise convergence), so continuity of $e$ gives $e(f(s), f_n(s)) = \lim_{m \to \infty} e(f_m(s), f_n(s)) \leq \varepsilon$. The weak inequality $\leq$ replaces $<$ because limits of strict inequalities yield non-strict ones. **From pointwise to uniform:** Since $e(f(s), f_n(s)) \leq \varepsilon$ holds for all $s \in S$, taking the supremum gives $\mathcal{D}(f_n, f) = \sup_{s \in S} e(f_n(s), f(s)) \leq \varepsilon$ for all $n \geq N$. The bound is independent of $s$ because the original Cauchy condition $\mathcal{D}(f_m, f_n) < \varepsilon$ was a sup-norm bound (uniform in $s$). This upgrade from pointwise to uniform convergence is the core content of the argument: the uniform Cauchy condition provides a bound that is preserved when passing to the limit. [/guided] [/step]