Metric spaces with countable dense subset Let $C^*(X)$ (endowed with sup norm) denote the metric space of all bounded real valued continuous functions on the metric space $X$. suppose that $C^*(X)$ contains a countable dense subset. I want to show that $X$ has a countable dense subset. Any idea or  reference is welcomed.   
 A: For each $n$, choose by Zorn's lemma a maximal $A_n\subseteq X$ such that the distance between any two points in $A_n$ is at least $1/n$. There are two cases:


*

*either each $A_n$ is countable, or

*some $A_n$ is uncountable.


In case 2. there is an uncountable family of disjoint open subsets of $X$. Using that you can construct such a family in $C^*(X)$, which would be a contradiction. In case 1., there is a countable dense subset of $X$.
A different flavour of (essentially) the same solution:


*

*Assume towards contradiction that $X$ does not have a countable dense subset and construct a sequence $x_\alpha$, $\alpha<\omega_1$ such that for each $\alpha$, the distance from $x_\alpha$ to any $x_\beta$ with $\beta<\alpha$ is at least $d_\alpha>0$.

*Now, there is some $n$ such that for uncountably many $\alpha$ we have $d_\alpha>1/n$. Let $X'$ be the set of just those $x_\alpha$.

*$X'$ is an uncountable set of points such that any two are at least $1/n$ apart.

*Now you can proceed as before.

A: I'm not sure the level of the answer that you are looking for but let me know if you want more details: If $C^*(X)$ is separable then so will be $C_b(X)$ the space of all bounded complex-valued functions. This is isomorphic to $C(\beta X)$, the algebra of all complex valued functions on the Stone-Čech compactification of $X$. Now for this to be separable implies that $\beta X$ is metrizable and this should imply that $\beta X=X$, thus $X$ is compact, and it is well known that every compact metric space is separable.
