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)
Is $BC([0,1))$ ( space of bounded real valued continuous functions) separable? Is $BC([0,1))$ a subset of $BC([0,\infty))$?
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2Let $\phi\colon [0,1) \to [0,\infty)$ be a homeomorphism. Then $\Phi\colon BC[0,\infty) \to BC[0,1)$, $f \mapsto f\circ \phi$ is an isomorphism. So $BC[0,1)$ is not separable as $BC[0,\infty)$ isn't. – 2012-03-12
2 Answers
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.
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0@t.b. Thanks. +1 to everyone! – 2012-03-12
$BC([0,1))$ is not a subset of $BC([0,\infty))$; in fact, these two sets of functions are disjoint. No function whose domain is $[0,1)$ has $[0,\infty)$ as its domain, and no function whose domain is $[0,\infty)$ has $[0,1)$ as its domains. What is true is that $\{f\upharpoonright[0,1):f\in BC([0,\infty))\}\subseteq BC([0,1))\;.$
There is, however, a very close relationship between $BC([0,\infty))$ and $BC([0,1))$, owing to the fact that $[0,\infty)$ and $[0,1)$ are homeomorphic. An explicit homeomorphism is $h:[0,\infty)\to[0,1):x\mapsto \frac2\pi\arctan x\;.$ This implies that $BC([0,1))$ and $BC([0,\infty))$ are actually homeomorphic, via the map $H:BC([0,1))\to BC([0,\infty)):f\mapsto f\circ h\;,$ as is quite easily checked. Thus, one of $BC([0,\infty))$ and $BC([0,1))$ is separable iff the other is.