It is easy to prove the non-separability of BC([0,$\infty$)) and the separability of C([0,1]). It seems to me we can argue from the fact that any bounded continuous function of BC([0,$\infty$)) must also be in BC([0,1)) to somehow show BC([0,1)) is not separable, but BC([0,1)
[Math] Is $BC([0,1))$ ( space of bounded real valued continuous functions) separable? Is $BC([0,1))$ a subset of $BC([0,\infty))$
analysisfunctional-analysisgeneral-topology
Best Answer
The space $BC([0,1))$ is not separable. For the begining consider function $$ \varphi(x)=\max(1-|2x|,0) $$ Then for each binary sequence $s\in\{0,1\}^\mathbb{N}$ we define function $$ g_s(x)=\sum\limits_{k=1}^\infty s(k)\varphi\left(x-\frac{k}{2}\right) $$ This is uncountable family in $BC([0,+\infty))$. Moreover if $s'\neq s''$, then $\Vert g_{s'}-g_{s''}\Vert_\infty=1$. Now consider functions $$ f_s(x)=g_s\left(\frac{1}{1-x}\right), \quad s\in\{0,1\}^\mathbb{N} $$ It is easy to see that $\{f_s:s\in\{0,1\}^\mathbb{N}\}$ is uncountable subset of $BC([0,1))$ and if $s'\neq s''$, then $\Vert f_{s'}-f_{s''}\Vert_\infty=1$. This is impossible if $BC([0,1))$ separable, so $BC([0,1))$ is not separable.