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Is it true, that if $X$ is a separable metric space, then the space of all continuous functions on $X$ with the supremum metric is also separable?

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I don't think it necessarily is. Consider the discrete metric space on $\Bbb N$. Then the set of bounded continuous functions is the space of bounded sequences $\ell^{\infty}$ which, if I remember correctly, is not separable. – Olivier Bégassat Feb 12 '13 at 11:48
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No. Consider triangle-shaped function $$ \varphi(x)=\max(1-2|x|,0) $$ then for each binary sequence $s:\mathbb{N}\to\{0,1\}$ we define $$ f_s(x)=\sum\limits_{n=1}^\infty s(n)\varphi(x-n) $$ One can show that $\{f_s:s\in\{0,1\}^\mathbb{N}\}$ is uncountable set of functions with the property $$ s'\neq s''\implies \Vert f_{s'}-f_{s''}\Vert_{C(\mathbb{R})}\geq 1 $$ This implies that $C_b(\mathbb{R})$ is not separable.

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Thanks very much, I'm asking because I read answer to this question… and it seems to me that the compactness is used only to show that the base for X is countable, so separability of X should be sufficient to prove my original proposition (which you disproved), what am I missing? – Josef Ondřej Feb 12 '13 at 12:20
@Josef: The linked post applies Stone-Weierstrass to deduce density. This theorem needs compactness for $C(X)$ or local compactness for the space $C_0(X)$ of functions vanishing at infinity. A further point you are missing is that the sup-metric is not really defined on all of $C(X)$, but on the bounded functions only. – Martin Feb 12 '13 at 13:16
@Martin: Aaaa, that explains it. Thanks very much. So in the case when X is compact, the proposition holds. I found it prooved also in a book by R. M. Dudley, but the proof seemed a bit more complicated. To the second remark, I think on a compact space all continuous functions must be bounded so I guess it shouldn't be a problem there, but you were right I was missing that in the more general case.… - Corollary 11.2.5 – Josef Ondřej Feb 12 '13 at 13:57

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