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Recall that $\ell_\infty$ is the space containing all bounded scalar-valued sequences with sup norm, that is, $$\|(x_n)_{n=1}^\infty\|_\infty = \sup_{n\in\mathbb{N}} |x_n|.$$

It is well-known that $\ell_\infty$ is not separable, that is, $\ell_\infty$ does not contain a countable dense subset. One way to show it is through establishing an injection from the set of all subsets of $\mathbb{N}$ into any given dense subset of $\ell_\infty.$

Let $BC(0,1)$ be a collection of all bounded scalar-valued continuous functions on $(0,1)$ with sup-norm $\|\cdot\|_\infty,$ that is, $$\|f\|_\infty = \sup_{t\in (0,1)}|f(t)|.$$

It can be shown that $\ell_\infty$ can be embedded isometrically onto a subspace of $C(0,1),$ through the mapping $$(x_n)_{n=1}^\infty \mapsto \sum_{n=1}^\infty x_nf_n$$ where $(f_n)_{n=1}^\infty$ is a collection of functions in $BC(0,1)$ with $\text{supp}(f_n)\subseteq (\frac{1}{n+1},\frac{1}{n})$ and $\|f_n\|_\infty = 1$ for all $n.$

Question: Can we conclude that $BC(0,1)$ is not separable?

Is it always true that when a normed space $(X,\|\cdot\|)$ contains an isometric copy of $\ell_\infty,$ $(X,\|\cdot\|)$ is not separable?

I think the answer is yes but I have no idea how to prove it.

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    $\begingroup$ I think this might help math.stackexchange.com/questions/516886/… If $X$ were separable, then $\ell^\infty$ would be separable, a contradiction. $\endgroup$ – Bowditch Apr 3 '18 at 11:43
  • $\begingroup$ Ooh yeah, you are right. I forgot that separability can be inherited as its is equivalent to second countability. $\endgroup$ – Idonknow Apr 3 '18 at 11:46
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    $\begingroup$ Just a word of caution, since I don't know how well you are aware of it: That's the case for metric spaces, but not for general topological spaces. In general, a separable space need not be second countable, and a subspace of a separable space need not be separable. $\endgroup$ – Daniel Fischer Apr 3 '18 at 11:56

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