How prove that $\mathcal{BC}([0,\infty))$ is not separable? How can we prove that the set $\mathcal{BC}([0,\infty))$ of bounded continuous functions on $[0,\infty)$, with the sup-norm, is not separable?
I'm currently struggling with the two following ideas: 


*

*Finding a surjection of $\mathcal{BC}([0,\infty))$ onto $l^\infty$, which is not separable (then the eventual countable, dense set of $\mathcal{BC}([0,\infty))$ would be mapped onto $l^\infty$, and this would be a contradiction)

*Using some separability result such as separability implies the existence of a countably orthonormal basis$\dots$  
 A: A continuous surjection onto $\ell^{\infty} $  is given by $ f\mapsto (f(1), f (2),\dots) $.
A: Your second idea won't work since $\mathcal {BC}([0,\infty))$ is not a Hilbert space. Your first idea is better, although I tend to find it easier to instead find an injective map $\ell^\infty\to\mathcal{BC}([0,\infty))$. But more direct is simply finding an uncountable family $\mathcal A\subset\mathcal{BC}([0,\infty))$ such that $\|f-g\|_\infty\ge\delta$ for all $f,g\in\mathcal A$, $f\neq g$, where $\delta>0$ is some fixed constant. Remember, $\mathcal{BC}([0,M])$ is separable for any $M>0$, so we need to find an example which takes advantage of the entire half line. Let $f_\lambda(x)=\sin(\lambda x)$; recall that if $\lambda/\pi$ is irrational, $\{f_\lambda(n)\,:\,n\in\mathbb N\}$ is dense in $[-1,1]$, so this seems like a good guess for an appropriate family of $\lambda$. Observe that, if $\epsilon=\mu-\lambda\neq0$,
$$|f_\lambda(x)-f_\mu(x)|=|\sin(\lambda x)-\sin(\lambda x)\cos(\epsilon x)-\cos(\lambda x)\sin(\epsilon x)|\\
\ge|\sin(\lambda x)|(1-\cos(\epsilon x))-|\cos(\lambda x)\sin(\epsilon x)|.$$
If we assume $\lambda/\epsilon$ is irrational, then there exists $k\in\mathbb N$ odd such that $|\sin(\frac\lambda\epsilon k\pi)|\ge\frac12$. So for $x=k\pi/\epsilon$, we have
$$|f_\lambda(x)-f_\mu(x)|\ge|\sin(\tfrac\lambda\epsilon k\pi)|\cdot2-0\ge1$$
and so $\|f_\lambda-f_\mu\|_\infty\ge1$.
Now let $B$ be a Hamel basis of $\mathbb R$ as a vector space over $\mathbb Q$, assuming without loss of generality that $\lambda>0$ for all $\lambda\in B$. Let $\mathcal A:=\{f_\lambda:\lambda\in B\}$. Then $\mathcal A$ is an uncountable family with the required property, so we are done.
A: Define $f_n(x) = \sin x\cdot \chi_{[n\pi,(n+1)\pi]}(x).$ Then $f_n \in \mathcal {BC}, n = 1,2,\dots.$ Let $A$ be the set of binary sequences. As is well known, $A$ is uncountable. For $a=(a_n)\in A,$ define
$$g(a) = \sum_{n=1}^{\infty}a_nf_n.$$
Then each $g(a) \in \mathcal {BC},$ and $a\ne b$ imples $\|g(a) - g(b)\|_{\mathcal {BC}}\ge 1.$ Thus $\mathcal {BC}$ contains an uncountable subset whose elements are all at distance at least one from each other. It follows that $\mathcal {BC}$ is not separable.
