Show that $(C_b (0,1],||\cdot||_{\infty})$ is not separable

Question

I'm not sure what I'm suppose to do for this question. Any help would be greatly appreciated. I know that there is a similar question like this for $[0,1)$ instead. But is there anyway for me to show this without having to prove the isomorphism between $[0,1)$ and $(0,1]$?

Answer 1

For any bounded sequence of real numbers $\mathfrak{a} =(a_n ) $ there exists a continuous function $f:(0,1]\to\mathbb{R}, $ such that $f_{\mathfrak{a}}(n^{-1} )=a_n$ and $\sup_n |a_n | =\sup_{\sigma\in (0,1]} |f_{\mathfrak{a}}(\sigma )|.$ Hence $\ell^{\infty}$ is isometricaly isomorphic to some subspace of $C_b (0,1].$ But $\ell^{\infty}$ is not separable and therefore $C_b (0,1]$ is not separable.

Answer 2

Hint: Think of triangles of height $1$ over the intervals $[1/(n+1),1/n], n = 1,2,\dots.$

Answer 3

More generally, if $X$ is normal and not limit-point compact, then $C_b(X)$ is not separable.

The essence of the argument is the same as the other answers: take $A$ an infinite subset which has no limit points. Then $A$ is closed on $X$, as also are all of its subsets. By Urysohn's lemma, for every $B \subset A$, there exists a continuous bounded function $f_B$ which is $1$ on $B$ and $0$ on $A-B$. It is clear that $\|f_B-f_{B'}\|_{\infty}=1$ for every $B \neq B'$. Since $A$ is infinite, $\mathcal{P}(A)$ is uncountable, and we've found an uncountable discrete subset of $C_b(X)$: namely, the set $\mathcal{F}:=\{f_B \mid B \in \mathcal{P}(A)\}$.

If $C_b(X)$ were separable, it would be second-countable (since it is metric), and therefore all of its subspaces would be second-countable. In particular, $\mathcal{F}$. But this is an absurd, since an uncountable discrete space cannot be secound-countable.