How to show that this is a complete metric space Let $(X;d)$ be a metric space and $\mathrm{C_b}(X,\mathbb{R})$ denote the set of all continuous bounded real valued functions defined on $X$, equipped with the uniform metric:
$$ \rho(f,g) = \sup\{\, \lvert f(x) - g(x)\lvert : x \in X \,\}. $$
Show that $\mathrm{C_b}(X,\mathbb{R})$ is a complete metric space with respect to the metric $\rho$.
 A: I leave it to you to make sure that $\rho$ is a metric. Note that if we have a sequence of functions $f_n\in Cb(X,\Bbb R)$ that is Cauchy with respect to $\rho$, then for each $x\in X$, the sequence $f_n(x)$ is Cauchy in $\Bbb R$, and so convergent. Define $f$ to be the pointwise limit of the $f_n$s. We must show that $f$ is bounded, continuous, and that $\rho(f_n,f)\to 0$ as $n\to\infty$. I leave boundedness to you.

Take $\epsilon>0$ and find $N$ such that $\rho(f_n,f_m)<\frac\epsilon2$ whenever $m,n\ge N$. Take any $n\ge N$, and take any $x\in X$. Since $f_m(x)$ converges to $f(x)$, then there is some $N'$ such that $|f_m(x)-f(x)|<\frac\epsilon2$ whenever $m\ge N'$. For $m\ge\max\{N,N'\},$ then, we have $$|f_n(x)-f(x)|\le|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$ Since $|f_n(x)-f(x)|<\epsilon$ for any $x\in X$, then $$\sup_{x\in X}|f_n(x)-f(x)|\le\epsilon.$$ Since this holds for all $n\ge N$, we have $$\lim_{n\to\infty}\sup_{x\in X}|f_n(x)-f(x)|=0.$$ If we can show that $f$ is continuous, then we will have $f\in Cb(X,\Bbb R)$ and $\rho(f_n,f)\to 0,$ as desired.

Take $\epsilon>0$, and find $N$ such that $\sup_{x\in X}|f_n(x)-f(x)|<\frac\epsilon3$ whenever $n\ge N$ (as we showed possible above). Take any $x_0\in X$. Since $f_N$ is continuous, then there exists $\delta>0$ such that $|f_N(y)-f_N(x_0)|<\frac\epsilon3$ whenever $y\in X$ with $d(y,x_0)<\delta$. Then for any $y\in X$ with $d(y,x_0)<\delta$ we have $$\begin{align}|f(y)-f(x_0)| &\le |f_N(y)-f(y)|+|f_N(y)-f_N(x_0)|+|f_N(x_0)-f(x_0)|\\ &\le |f_N(y)-f_N(x_0)|+2\sup_{x\in X}|f_N(x)-f(x)|\\ &< \frac\epsilon3+\frac{2\epsilon}3\\ &=\epsilon,\end{align}$$ so $f$ is continuous at $x_0$. Since $x_0\in X$ was arbitrary, then $f$ is continuous, and so we're done.
