# Separability of $B(X)$.

Assume $$X$$ is a separable space, or even a countable one. Is then the space of bounded complex-valued functions on $$X$$ with the supremum norm, $$\|f\|_{\infty}=\sup_{x\in X} |f(x)|$$, separable as well?

Edit: Assume that $$X$$ is a countable set. Is the answer still no? All the counterexamples seem to consider $$X$$ as uncountable.

• Consider indicator functions of countable sets. Commented Jul 19, 2023 at 18:00
• Commented Jul 19, 2023 at 18:08
• $\ell^\infty(\Bbb N)$ is also not separable. Commented Jul 19, 2023 at 19:26

The space $$\ell^\infty(\mathbb N)$$ is not separable. An easy way to see this is that it contains all sequences of zeros and ones (which can be identified with the power set of $$\mathbb N$$), and the distance between any two of these sequences is $$1$$.
• ... and these sequences have distance $1$ from each other, otherwise this would not rule out separability. Commented Jul 20, 2023 at 8:34
The Banach space $$L^{\infty }(\mathbb{R})$$ is not separable, even though $$\mathbb{R}$$ is.