[proofplan] Start with an arbitrary [Cauchy sequence](/page/Cauchy%20Sequence) $(f_n)_{n=1}^{\infty}$ in the sup norm. Evaluating at a fixed point $x\in X$ turns the sequence into a Cauchy sequence of [real numbers](/page/Real%20Numbers), so completeness of $\mathbb{R}$ gives a pointwise candidate limit $f:X\to\mathbb{R}$. The same sup-norm Cauchy estimate, after passing to the pointwise limit in one index, gives [uniform convergence](/page/Uniform%20Convergence) $f_n\to f$. Finally, boundedness follows by comparing $f$ to one bounded tail function, and continuity follows directly from uniform convergence and continuity of that same tail function. [/proofplan]

[step:Construct the pointwise candidate limit from real completeness]Let $(f_n)_{n=1}^{\infty}$ be a Cauchy sequence in $(C_b(X),\|\cdot\|_\infty)$. Thus, for every $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that for all $m,n\geq N$, \begin{align*} \|f_n-f_m\|_\infty<\varepsilon. \end{align*} Fix $x\in X$. Since \begin{align*} |f_n(x)-f_m(x)|\leq \|f_n-f_m\|_\infty \end{align*} for all $m,n\in\mathbb{N}$, the scalar sequence $(f_n(x))_{n=1}^{\infty}$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, it converges to a real number. Define the function \begin{align*} f:X\to\mathbb{R} \end{align*} by requiring, for each $x\in X$, \begin{align*} f(x)=\lim_{n\to\infty}f_n(x). \end{align*}[/step]

[guided]We begin with a Cauchy sequence $(f_n)_{n=1}^{\infty}$ in the normed space $(C_b(X),\|\cdot\|_\infty)$. By definition, this means that for every $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that whenever $m,n\geq N$, \begin{align*} \|f_n-f_m\|_\infty<\varepsilon. \end{align*} The first task is to produce a plausible limit function. Since functions are determined by their values at points, fix an arbitrary point $x\in X$. The sup norm controls every pointwise difference, because the supremum of $|f_n(y)-f_m(y)|$ over all $y\in X$ is at least the value at the particular point $x$. Therefore, \begin{align*} |f_n(x)-f_m(x)|\leq \|f_n-f_m\|_\infty. \end{align*} It follows that $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$. The real numbers are complete, so this scalar sequence has a real limit. Because this construction works for every $x\in X$, we define a function \begin{align*} f:X\to\mathbb{R} \end{align*} by setting \begin{align*} f(x)=\lim_{n\to\infty}f_n(x) \end{align*} for each $x\in X$. This gives the only possible pointwise limit of the sequence; the remaining work is to prove that the convergence is actually in the sup norm and that $f$ belongs to $C_b(X)$.[/guided]

[step:Upgrade the Cauchy estimate to uniform convergence] Let $\varepsilon>0$. Since $(f_n)_{n=1}^{\infty}$ is Cauchy in $\|\cdot\|_\infty$, choose $N\in\mathbb{N}$ such that for all $m,n\geq N$, \begin{align*} \|f_n-f_m\|_\infty<\frac{\varepsilon}{2}. \end{align*} Fix $n\geq N$ and $x\in X$. Then, for every $m\geq N$, \begin{align*} |f_n(x)-f_m(x)|<\frac{\varepsilon}{2}. \end{align*} Passing to the limit as $m\to\infty$ in the real-valued sequence $(f_m(x))_{m=1}^{\infty}$ gives \begin{align*} |f_n(x)-f(x)|\leq \frac{\varepsilon}{2}<\varepsilon. \end{align*} Since this estimate holds for every $x\in X$, we have \begin{align*} \|f_n-f\|_\infty\leq \frac{\varepsilon}{2}<\varepsilon. \end{align*} Therefore $f_n\to f$ uniformly on $X$, equivalently $f_n\to f$ in the sup norm once $f$ is known to be bounded. [/step]

[step:Prove that the pointwise limit is bounded] Apply the uniform convergence just proved with $\varepsilon=1$. There exists $N\in\mathbb{N}$ such that \begin{align*} \|f_N-f\|_\infty<1. \end{align*} Because $f_N\in C_b(X)$, it is bounded, so the real number $\|f_N\|_\infty$ is finite. For every $x\in X$, the triangle inequality in $\mathbb{R}$ gives \begin{align*} |f(x)|\leq |f(x)-f_N(x)|+|f_N(x)|. \end{align*} Using the definitions of the sup norm terms, \begin{align*} |f(x)|\leq \|f-f_N\|_\infty+\|f_N\|_\infty<1+\|f_N\|_\infty. \end{align*} Thus $f:X\to\mathbb{R}$ is bounded. [/step]

[step:Prove continuity of the pointwise limit directly from uniform convergence]Fix $x_0\in X$ and let $O\subset\mathbb{R}$ be an [open set](/page/Open%20Set) with $f(x_0)\in O$. Since $O$ is open in $\mathbb{R}$, choose $\eta>0$ such that \begin{align*} (f(x_0)-\eta,f(x_0)+\eta)\subset O. \end{align*} By uniform convergence, choose $N\in\mathbb{N}$ such that \begin{align*} \|f_N-f\|_\infty<\frac{\eta}{3}. \end{align*} Since $f_N:X\to\mathbb{R}$ is continuous at $x_0$, there exists an open neighbourhood $U\subset X$ of $x_0$ such that for every $x\in U$, \begin{align*} |f_N(x)-f_N(x_0)|<\frac{\eta}{3}. \end{align*} For every $x\in U$, the triangle inequality gives \begin{align*} |f(x)-f(x_0)|\leq |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|. \end{align*} Each of the first and third terms is bounded by $\|f-f_N\|_\infty$, and the middle term is bounded by the choice of $U$. Hence \begin{align*} |f(x)-f(x_0)|<\eta. \end{align*} Therefore $f(U)\subset O$. Since $x_0\in X$ and the open neighbourhood $O$ of $f(x_0)$ were arbitrary, $f:X\to\mathbb{R}$ is continuous.[/step]

[guided]The point of this step is to prove continuity without relying on a separately cited [uniform limit theorem](/theorems/258). Fix a point $x_0\in X$ and an open set $O\subset\mathbb{R}$ containing $f(x_0)$. To show continuity at $x_0$, we must find an open neighbourhood $U\subset X$ of $x_0$ such that $f(U)\subset O$. Because $O$ is open in the usual topology on $\mathbb{R}$ and contains $f(x_0)$, there is a radius $\eta>0$ such that \begin{align*} (f(x_0)-\eta,f(x_0)+\eta)\subset O. \end{align*} The strategy is to compare $f$ to one [continuous function](/page/Continuous%20Function) $f_N$. Uniform convergence lets us choose this single index $N$ so that the approximation is good at every point of $X$: \begin{align*} \|f_N-f\|_\infty<\frac{\eta}{3}. \end{align*} Now use the continuity of $f_N:X\to\mathbb{R}$ at $x_0$. There exists an open neighbourhood $U\subset X$ of $x_0$ such that for every $x\in U$, \begin{align*} |f_N(x)-f_N(x_0)|<\frac{\eta}{3}. \end{align*} For such an $x$, insert and subtract the two approximating values $f_N(x)$ and $f_N(x_0)$. The triangle inequality in $\mathbb{R}$ gives \begin{align*} |f(x)-f(x_0)|\leq |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|. \end{align*} The first and third terms are controlled by the uniform estimate $\|f_N-f\|_\infty<\eta/3$, while the middle term is controlled by the continuity of $f_N$ on the neighbourhood $U$. Therefore, \begin{align*} |f(x)-f(x_0)|<\frac{\eta}{3}+\frac{\eta}{3}+\frac{\eta}{3}=\eta. \end{align*} Thus $f(x)\in (f(x_0)-\eta,f(x_0)+\eta)\subset O$ for every $x\in U$, so $f(U)\subset O$. This proves continuity of $f$ at $x_0$, and since $x_0$ was arbitrary, $f:X\to\mathbb{R}$ is continuous.[/guided]

[step:Conclude completeness of the normed space] The previous steps show that the pointwise limit $f:X\to\mathbb{R}$ is bounded and continuous, so $f\in C_b(X)$. The uniform convergence estimate gives \begin{align*} \lim_{n\to\infty}\|f_n-f\|_\infty=0. \end{align*} Thus every Cauchy sequence in $(C_b(X),\|\cdot\|_\infty)$ converges in the sup norm to an element of $C_b(X)$. Therefore $(C_b(X),\|\cdot\|_\infty)$ is complete. [/step]