Let $X$ be a complete and separable metric space. Define
\begin{equation} M := \{ f \, \mid \, f : X \rightarrow \mathbb{R}, \, f \text{ bounded and continuous } \} \end{equation}
Is M separable with respect to supremum norm?
Let $X$ be a complete and separable metric space. Define
\begin{equation} M := \{ f \, \mid \, f : X \rightarrow \mathbb{R}, \, f \text{ bounded and continuous } \} \end{equation}
Is M separable with respect to supremum norm?
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No, not necessarily: take $X= \mathbb{N}$ in the discrete metric, which is complete and separable and then $M = \ell^\infty$ which is not separable.
Stone-Weierstrass says yes for $X$ compact metric, e.g.