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.
Countable total subset in $\ell^\infty/ c_0$
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functional-analysis
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0The symbol '\' is restricted, and so I used '\backslash' instead. – 2012-09-24
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0You 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". – 2012-09-24