Is $(\mathcal C_b(\mathbb R),\|\cdot \|_\infty )$ space of bounded continuous function complete?

Question

Let $\mathcal C_b(\mathbb R)$ the space of bounded continuous function on $\mathbb R$ endowed with the sup norm $\|\cdot \|_\infty $. Is this space complete ? So, let $(f_n)$ a Cauchy sequence. In particular, $(f_n(x))$ is Cauchy as well and thus converges to some $f(x)$.

The proof should be the same than my proof of $(\mathcal C[0,1],\|\cdot \|_\infty )$ complete here, but unfortunately I can't prove the fact that $\|f_n-f\|_\infty $ is finite (I can do it in $(\mathcal C[0,1],\|\cdot \|_\infty )$, but I can't adapt it in $(\mathcal C_b(\mathbb R),\|\cdot \|_\infty )$, since in this space I can't use Bolzano-weierstrass). In $\mathcal C([0,1])$ the proof goes as follow : suppose $\|f-f_n\|_\infty =\infty $, i.e. for all $m$, there is $x_m^n\in [0,1]$ s.t. $|f_n(x_m^n)-f(x_m^n)|\geq m$. Using Bolzano-Weierstass, there is a sub-sequence still denoted $(x_m^n)$ that converges to $x\in [0,1]$. Therefore, $$0=\lim_{n\to \infty }|f_n(x_m^n)-f(x_m^n)|\geq m,$$ which is a contradicton. But this doesn't work if the sequence leaves in $\mathbb R$ instead of $[0,1]$.

Answer 1

$(f_n)$ is cauchy, so there exists a $n \in \mathbb{N}$ such that $||f_n - f_m|| < 1$ for all $m > n$. Therefore, we get $|f_n(x) - f_m(x)| < 1$ for all $m > n$ and $x \in \mathbb{R}$. By taking the limit $m \to \infty$ we get $|f_n(x) - f(x)| \leq 1$ for all $x \in \mathbb{R}$. So we get that $||f_n - f|| \leq 1$ which means that in particular $f$ is again bounded and therefore $||f_n - f|| < \infty$ for all $n \in \mathbb{N}$.