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  "body": "\n## Proof\n\nWe first check that $d_\\infty$ is a metric. Then we build the limit of a Cauchy sequence point by point and show it is uniform.\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]] and let $(Y, d_Y)$ be a [[complete metric space|08EWT7J0SC]]. Write $C_b = C_b(X, Y)$.\n2. If $X$ is empty, then $C_b$ has one element, the empty map. Then $d_\\infty$ is a [[metric|063W5JMY85]], and each sequence is constant and converges. So we assume $X \\ne \\emptyset$ from now on.\n3. Let $f, g \\in C_b$ and fix $x_0 \\in X$. Since $f(X)$ and $g(X)$ are [[bounded|0P953MG05E]], there are reals $M_f$ and $M_g$ with $d_Y(f(x), f(x')) \\le M_f$ and $d_Y(g(x), g(x')) \\le M_g$ for all $x, x' \\in X$.\n4. By the [[triangle inequality|01M6CW9VM6]], $d_Y(f(x), g(x)) \\le d_Y(f(x), f(x_0)) + d_Y(f(x_0), g(x_0)) + d_Y(g(x_0), g(x)) \\le M_f + d_Y(f(x_0), g(x_0)) + M_g$ for each $x \\in X$.\n5. So the set $\\set{d_Y(f(x), g(x)) : x \\in X}$ is nonempty and bounded above. Its [[supremum|0BTMQ04556]] $d_\\infty(f, g)$ is a real number $\\ge 0$.\n6. We have $d_\\infty(f, f) = 0$ and $d_\\infty(f, g) = d_\\infty(g, f)$, since $d_Y$ has these properties at each point.\n7. If $f \\ne g$, there is $x$ with $f(x) \\ne g(x)$. Then $d_\\infty(f, g) \\ge d_Y(f(x), g(x)) \u003e 0$.\n8. Let $h \\in C_b$. For each $x$, $d_Y(f(x), h(x)) \\le d_Y(f(x), g(x)) + d_Y(g(x), h(x)) \\le d_\\infty(f, g) + d_\\infty(g, h)$. Taking the supremum over $x$ gives the triangle inequality for $d_\\infty$.\n9. By steps 5 to 8, $d_\\infty$ is a metric on $C_b$.\n10. Let $(f_n)$ be a [[Cauchy sequence|0MCK8K2D7Z]] in $(C_b, d_\\infty)$. Fix $x \\in X$.\n11. For all $m, n$, $d_Y(f_m(x), f_n(x)) \\le d_\\infty(f_m, f_n)$. So $(f_n(x))$ is a Cauchy sequence in $Y$.\n12. Since $Y$ is complete, $(f_n(x))$ [[converges|06JYPJ4WHH]] to a point of $Y$. Call it $f(x)$. This defines a map $f \\colon X \\to Y$.\n13. Let $\\varepsilon \u003e 0$. Choose $N$ with $d_\\infty(f_m, f_n) \u003c \\varepsilon$ for all $m, n \\ge N$.\n14. Fix $x \\in X$ and $n \\ge N$. Since $f_m(x) \\to f(x)$, there is $m \\ge N$ with $d_Y(f_m(x), f(x)) \u003c \\varepsilon$.\n15. Then $d_Y(f_n(x), f(x)) \\le d_Y(f_n(x), f_m(x)) + d_Y(f_m(x), f(x)) \u003c 2 \\varepsilon$.\n16. So, for each $\\varepsilon \u003e 0$, there is $N$ with $d_Y(f_n(x), f(x)) \u003c 2 \\varepsilon$ for all $n \\ge N$ and all $x \\in X$. Call this fact (U).\n17. The image $f(X)$ is bounded. Apply (U) with $\\varepsilon = 1$ and take $n = N$. For $x, x' \\in X$, the triangle inequality gives $d_Y(f(x), f(x')) \u003c 2 + M_{f_n} + 2$.\n18. Next, $f$ is [[continuous|0SY9ECQ19Q]]. Let $W$ be open in $Y$ and let $x \\in f^{-1}(W)$.\n19. By the description of [[open sets of a metric space|0A26X79RTB]], there is $r \u003e 0$ with $B(f(x), r) \\subseteq W$. Here $B$ is the [[open ball|0DC3MPR7C9]].\n20. Apply (U) with $\\varepsilon = r / 6$. This gives $n$ with $d_Y(f_n(z), f(z)) \u003c r / 3$ for all $z \\in X$.\n21. Let $V = f_n^{-1}(B(f_n(x), r / 3))$. [[Open balls are open|0B2305GNAZ]] and $f_n$ is continuous, so $V$ is open in $X$. Also $x \\in V$.\n22. Let $y \\in V$. Then $d_Y(f(y), f(x)) \\le d_Y(f(y), f_n(y)) + d_Y(f_n(y), f_n(x)) + d_Y(f_n(x), f(x)) \u003c r$. So $f(y) \\in W$.\n23. So each point $x$ of $f^{-1}(W)$ lies in an open set $V$ with $V \\subseteq f^{-1}(W)$. Thus $f^{-1}(W)$ is the union of these open sets, and it is open.\n24. By steps 17 to 23, $f \\in C_b$.\n25. By (U), for each $\\varepsilon \u003e 0$ there is $N$ with $d_\\infty(f_n, f) \\le 2 \\varepsilon$ for all $n \\ge N$. So $f_n \\to f$ in $(C_b, d_\\infty)$.\n26. Thus each Cauchy sequence in $C_b$ converges in $C_b$, and $(C_b, d_\\infty)$ is a complete metric space.\n\n## Idea\n\nA Cauchy sequence for the sup metric is Cauchy at each point, so completeness of $Y$ gives a pointwise limit.\nThe Cauchy bound holds for all points at once, so the convergence is uniform. A uniform limit keeps boundedness and continuity, by an $\\varepsilon / 3$ argument.\n",
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