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In the post When is $C_0(X)$ separable? , it is argued that if $X$ is second countable and locally compact Hausdorff, then $C_{0}(X)$ is separable. Is it also true that $C(X)$ is separable under the same hypotheses?

Community
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cyc
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  • What topology are you considering on $C(X)$? Unless $X$ is compact, the sup norm will not make sense -- you would need to restrict to the set $C_b(X)$ of bounded continuous functions on $X$. – Scott LaLonde Apr 28 '13 at 03:25
  • Does it make sense to consider the topology of uniform convergence, i.e. the topology generated by the sets ${g:X\rightarrow\mathbb{C}|\sup_{x\in X}|g(x)-f(x)|<n^{-1}}$ where $n\in\mathbb{N},f:X\rightarrow\mathbb{C}$? – cyc Apr 28 '13 at 03:36
  • It makes sense to consider the topology of uniform convergence. However, I think that topology is "too big" to be separable in general, even when $X$ is second countable and LCH. – Scott LaLonde Apr 28 '13 at 04:23

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It is not true in general. For example $C(\mathbb{R})$ is not separable because you can find uncountable familily of continuous functions with pairwise distance equal to $1$. Below I present its simple counstrucion. Take arbitrary binary sequence $w\in\{0,1\}^\mathbb{N}$ and consider continuous function $$ f_w(x)=\sum\limits_{k=1}^\infty w_k\max(1-|2x-k|,0) $$ It consists of triangle shaped peaks at point $k/2$ if $w_k=1$ and plato plateau if $w_k=0$. The desired family of functions is $\{f_w:w\in\{0,1\}^\mathbb{N}\}$.

Norbert
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It seems that even $C(\mathbb R)$ or ($C(\mathbb Z)$) in the topology of uniform convergence is not separable. Suppose the opposite. Let $\{f_n:n\in\mathbb N\}$ be a separable subset of $C(\mathbb R)$. A function $f’:\mathbb N\to\mathbb R$ such that $f’(n)=f_n(n)+1$ is continuous on a discrete space $\mathbb N$. Tietze-Urysohn theorem implies that there exists a continuous extension $f:\mathbb R\to\mathbb R$ of the function $f’$. Then $||f-f_n||\ge 1$ for each $n$, a contradiction.

Alex Ravsky
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