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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.

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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 $1$.

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  • $\begingroup$ ... and these sequences have distance $1$ from each other, otherwise this would not rule out separability. $\endgroup$
    – MaoWao
    Commented Jul 20, 2023 at 8:34
  • $\begingroup$ Indeed, I have edited that in. Thanks! $\endgroup$ Commented Jul 20, 2023 at 8:44
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The Banach space $L^{\infty }(\mathbb{R})$ is not separable, even though $\mathbb{R}$ is.

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