[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*}
[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]
[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.
[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]
[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]
