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

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 $$[0,1]$$.

$$(f_n)$$ 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 $$n \in \mathbb{N}$$.
• Very Very Very nice ! (and elegant) You are a king. Thank you :) So at the end $(\mathcal C_b(\mathbb R),\|\cdot \|_\infty )$ is complete, right ?