TITLE: The space of bounded continuous functions are not separable
QUESTION [45 upvotes]: Let $C_b(\mathbb{R})$ be the space of all bounded continuous functions on $\mathbb{R}$, normed with 
$$\|f\| = \sup_{x\in \mathbb{R}}{\lvert f(x)\rvert}$$
Show that the space $C_b(\mathbb{R})$ is not separable.
A space is separable if there is a dense countable subset.
How do I prove that something is not separable? would it matter if we looked at $C_b([0,1])$ instead?

REPLY [17 votes]: Here is also one way, incase you know that $\ell^{\infty}$ is not separable, the set of bounded sequences with sup-norm that is.
Define $\Lambda:C_{b}(\mathbb{R})\to\ell^{\infty}$ by $f\mapsto (f(n))_{n=1}^{\infty}$. By choosing any $(x_{n})_{n=1}^{\infty}\in \ell^{\infty}$ we can define a continuous bounded function $f\in C_{b}(\mathbb{R})$ by setting $f(n)=x_{n}$ for all $n\in\mathbb{N}$ and extending it to $\mathbb{R}$ by connecting the imagepoints of each natural number linearly. We may take $f(x)= x_{1}$ for all $x<1$, for example. Since $f\mapsto (x_{n})_{n=1}^{\infty}$, then $\Lambda$ is a surjection. It is also continuous, because
\begin{equation*}
\|\Lambda(f)-\Lambda(g)\|_{\infty}=\sup_{n\in\mathbb{N}}|f(n)-g(n)|\leq \sup_{x\in\mathbb{R}}|f(x)-g(x)|=\|f-g\|_{\infty}
\end{equation*}
for all $f,g\in C_{b}(\mathbb{R})$.
Now if $C_{b}(\mathbb{R})$ was separable, it would have a countable and dense subset $\mathscr{D}$. Since the image of a dense subset under a continuous surjection is dense, we have that $\Lambda(\mathscr{D})$ is a countable, dense subset of $\ell^{\infty}$. This would imply that $\ell^{\infty}$ is separable, which is a contradiction.