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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)

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Let $\phi\colon [0,1) \to [0,\infty)$ 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. –  martini Mar 12 '12 at 8:49

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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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Thanks. But I can't accept or thumb up your answer because my reputation is less than 15. I will do it when I can. –  Polymorpher Mar 12 '12 at 12:23
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Glad that my answer helped you. –  Norbert Mar 12 '12 at 12:47
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@Polymorpher: now you can :) –  t.b. Mar 12 '12 at 14:27
    
@t.b. Thanks. +1 to everyone! –  Polymorpher Mar 12 '12 at 15:11

$BC([0,1))$ 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.

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