[proofplan] We prove completeness directly. A $d_\infty$-[Cauchy sequence](/page/Cauchy%20Sequence) of bounded continuous maps is pointwise Cauchy, so completeness of $Y$ gives a pointwise limit map $f:X\to Y$. The uniform Cauchy condition then upgrades pointwise convergence to [uniform convergence](/page/Uniform%20Convergence), and the uniform limit is shown to be bounded and continuous by elementary triangle-inequality estimates. Hence every Cauchy sequence in $C_b(X,Y)$ converges in $C_b(X,Y)$. [/proofplan]

[step:Handle the empty domain separately] If $X=\varnothing$, there is exactly one map $\varnothing\to Y$. Hence $C_b(\varnothing,Y)$ is a one-point [metric space](/page/Metric%20Space), and every Cauchy sequence converges to its unique element. Therefore $(C_b(\varnothing,Y),d_\infty)$ is complete. For the rest of the proof, assume $X\neq\varnothing$. [/step]

[step:Construct the pointwise limit from a uniformly Cauchy sequence] Let $(f_n)_{n\in\mathbb N}$ be a $d_\infty$-Cauchy sequence in $C_b(X,Y)$. Thus each $f_n:X\to Y$ is continuous and bounded, and for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for all $m,n\geq N$, \begin{align*} d_\infty(f_m,f_n)<\varepsilon. \end{align*} Fix $x\in X$. Since \begin{align*} e(f_m(x),f_n(x))\leq d_\infty(f_m,f_n) \end{align*} for all $m,n\in\mathbb N$, the sequence $(f_n(x))_{n\in\mathbb N}$ is Cauchy in $(Y,e)$. Because $(Y,e)$ is complete, there exists a unique point of $Y$, denoted $f(x)$, such that \begin{align*} \lim_{n\to\infty} e(f_n(x),f(x))=0. \end{align*} This defines a map $f:X\to Y$ by assigning to each $x\in X$ the limit of $(f_n(x))_{n\in\mathbb N}$. [/step]

[step:Upgrade pointwise convergence to convergence in the uniform metric]We prove that $f_n\to f$ uniformly. Let $\varepsilon>0$. Since $(f_n)_{n\in\mathbb N}$ is $d_\infty$-Cauchy, choose $N\in\mathbb N$ such that for all $m,n\geq N$, \begin{align*} d_\infty(f_m,f_n)<\varepsilon. \end{align*} Fix $n\geq N$ and $x\in X$. For every $m\geq N$, \begin{align*} e(f_n(x),f_m(x))\leq d_\infty(f_n,f_m)<\varepsilon. \end{align*} For every $m\geq N$, the triangle inequality gives \begin{align*} e(f_n(x),f(x))\leq e(f_n(x),f_m(x))+e(f_m(x),f(x))<\varepsilon+e(f_m(x),f(x)). \end{align*} Letting $m\to\infty$ and using $f_m(x)\to f(x)$ in $(Y,e)$ gives \begin{align*} e(f_n(x),f(x))\leq\varepsilon. \end{align*} Since this bound holds for every $x\in X$, \begin{align*} d_\infty(f_n,f)\leq\varepsilon \end{align*} for every $n\geq N$. Hence $d_\infty(f_n,f)\to 0$ as $n\to\infty$.[/step]

[guided]The Cauchy condition controls all tails of the sequence uniformly in $x$. We want to pass from the estimate between two sequence terms, $f_n$ and $f_m$, to an estimate between $f_n$ and the limiting map $f$. Let $\varepsilon>0$. Because $(f_n)_{n\in\mathbb N}$ is Cauchy in the metric $d_\infty$, there exists $N\in\mathbb N$ such that whenever $m,n\geq N$, \begin{align*} d_\infty(f_m,f_n)<\varepsilon. \end{align*} By the definition of $d_\infty$, this means that for every $x\in X$ and all $m,n\geq N$, \begin{align*} e(f_m(x),f_n(x))<\varepsilon. \end{align*} Now fix $n\geq N$ and $x\in X$. The point $f(x)\in Y$ was defined as the limit of the sequence $(f_m(x))_{m\in\mathbb N}$. For every $m\geq N$, the triangle inequality gives \begin{align*} e(f_n(x),f(x))\leq e(f_n(x),f_m(x))+e(f_m(x),f(x))<\varepsilon+e(f_m(x),f(x)). \end{align*} Since $f_m(x)\to f(x)$ in the metric space $(Y,e)$, we have $e(f_m(x),f(x))\to 0$. Letting $m\to\infty$ in the preceding inequality gives \begin{align*} e(f_n(x),f(x))\leq\varepsilon. \end{align*} The important point is that $N$ does not depend on $x$. Taking the supremum over all $x\in X$ gives \begin{align*} d_\infty(f_n,f)=\sup_{x\in X} e(f_n(x),f(x))\leq\varepsilon \end{align*} for every $n\geq N$. This is exactly convergence of $f_n$ to $f$ in the uniform metric.[/guided]

[step:Show the uniform limit is bounded] Choose $N\in\mathbb N$ such that \begin{align*} d_\infty(f_N,f)\leq 1. \end{align*} Since $f_N\in C_b(X,Y)$, its image has finite diameter. Define \begin{align*} M_N:=\sup_{x,y\in X} e(f_N(x),f_N(y))<\infty. \end{align*} For arbitrary $x,y\in X$, the triangle inequality in $(Y,e)$ gives \begin{align*} e(f(x),f(y))\leq e(f(x),f_N(x))+e(f_N(x),f_N(y))+e(f_N(y),f(y)). \end{align*} Using $d_\infty(f_N,f)\leq 1$ and the definition of $M_N$, we obtain \begin{align*} e(f(x),f(y))\leq 1+M_N+1=M_N+2. \end{align*} Taking the supremum over $x,y\in X$ shows that $f(X)$ has finite diameter. Thus $f:X\to Y$ is bounded. [/step]

[step:Prove continuity of the uniform limit]Let $x_0\in X$ and let $\varepsilon>0$. Choose $N\in\mathbb N$ such that \begin{align*} d_\infty(f_N,f)<\frac{\varepsilon}{3}. \end{align*} The map $f_N:X\to Y$ is continuous at $x_0$. Therefore there exists an open neighbourhood $U\subset X$ of $x_0$ such that for every $x\in U$, \begin{align*} e(f_N(x),f_N(x_0))<\frac{\varepsilon}{3}. \end{align*} For every $x\in U$, the triangle inequality gives \begin{align*} e(f(x),f(x_0))\leq e(f(x),f_N(x))+e(f_N(x),f_N(x_0))+e(f_N(x_0),f(x_0)). \end{align*} Each outer term is bounded by $d_\infty(f_N,f)$, and the middle term is bounded by the choice of $U$. Hence \begin{align*} e(f(x),f(x_0))<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon. \end{align*} Thus $f$ is continuous at $x_0$. Since $x_0\in X$ was arbitrary, $f:X\to Y$ is continuous.[/step]

[guided]We know that $f$ is uniformly close to some [continuous function](/page/Continuous%20Function) $f_N$. To prove continuity of $f$ at a point $x_0\in X$, we compare $f(x)$ to $f(x_0)$ by inserting the intermediate values $f_N(x)$ and $f_N(x_0)$. Fix $x_0\in X$ and $\varepsilon>0$. Since $f_n\to f$ in the uniform metric, choose $N\in\mathbb N$ such that \begin{align*} d_\infty(f_N,f)<\frac{\varepsilon}{3}. \end{align*} This uniform estimate means that for every $z\in X$, \begin{align*} e(f_N(z),f(z))<\frac{\varepsilon}{3}. \end{align*} Now use the continuity of the single map $f_N:X\to Y$ at $x_0$. There exists an open neighbourhood $U\subset X$ of $x_0$ such that for all $x\in U$, \begin{align*} e(f_N(x),f_N(x_0))<\frac{\varepsilon}{3}. \end{align*} For such an $x$, the triangle inequality gives \begin{align*} e(f(x),f(x_0))\leq e(f(x),f_N(x))+e(f_N(x),f_N(x_0))+e(f_N(x_0),f(x_0)). \end{align*} The first and third terms are controlled by the uniform closeness of $f_N$ to $f$, while the middle term is controlled by continuity of $f_N$ at $x_0$. Therefore \begin{align*} e(f(x),f(x_0))<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon. \end{align*} This proves continuity of $f$ at $x_0$. Since $x_0$ was arbitrary, $f:X\to Y$ is continuous on $X$.[/guided]

[step:Conclude that every Cauchy sequence converges in $C_b(X,Y)$] The preceding steps show that the $d_\infty$-Cauchy sequence $(f_n)_{n\in\mathbb N}$ converges in the uniform metric to a map $f:X\to Y$ that is both bounded and continuous. Hence $f\in C_b(X,Y)$ and \begin{align*} \lim_{n\to\infty} d_\infty(f_n,f)=0. \end{align*} Therefore every Cauchy sequence in $(C_b(X,Y),d_\infty)$ converges to an element of $C_b(X,Y)$. This proves that $(C_b(X,Y),d_\infty)$ is complete. [/step]