In an answer for the question: Complement of $c_{0}$ in $\ell^{\infty}$ the author argues that $\ell_\infty/ c_0$ has no countable total subset while $\ell_\infty$ does. It wasn't clear for me why this is true. Any clarification is appreciated.
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The symbol '' is restricted, and so I used '\backslash' instead. – M Turgeon Sep 24 '12 at 14:26
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1You seem to be referring to t.b.'s comment on GEdgar's answer. But what t.b. actually says is that "the dual of $\ell^\infty$ contains a countable total subset, while the dual of $\ell^\infty / c_0$ does not". – Chris Eagle Sep 24 '12 at 14:35
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Let $\{A_\alpha:\alpha\in \mathcal{A}\}$ be an uncountable family of almost disjoint subsets of $\mathbb{N}$. For construction of these sets see this answer.
For each $\alpha\in \mathcal{A}$ by $x_\alpha$ we denote coset in $\ell_\infty/c_0$ of the characteristic function of $A_\alpha$. Using ideas of section 2 of this answer you can show that $\{x_\alpha\in\mathcal{A}:g(x_\alpha)\neq 0 \}$ is at most countable for any $g\in(\ell_\infty/c_0)$. This means that for any countable family $\{h_k:k\in\mathbb{N}\}\subset (\ell_\infty/c_0)^*$ there are only countably many $x_\alpha\in \ell_\infty/c_0$ such that $h_k(x_\alpha)$ is non zero for some $k\in\mathbb{N}$. Hence we can find $x_{\alpha_0}$ which is mapped to zero by $h_k$ for all $k\in\mathbb{N}$. So $\{h_k:k\in\mathbb{N}\}$ is no total subset. Thus $(\ell_\infty/c_0)^*$ does not have countable total subset.
Meanwhile $(\ell_\infty)^*$ does have such a subset. Just consider functionals $f_k:\ell_\infty\to\mathbb{C}:x\mapsto x(k)$ where $k\in\mathbb{N}$.
