Is the Banach space \(\ell^\infty\) of bounded complex sequences, with the supremum norm, separable? Prove your answer.
Solution
No. The space \(\ell^\infty\) is not separable.
For each subset \(A\subset\mathbb N\), let \(x^A\in\ell^\infty\) be its characteristic sequence: \[ x^A_n=\begin{cases}1,&n\in A,\\0,&n\notin A.\end{cases} \] If \(A\ne B\), then for some \(n\) exactly one of \(A,B\) contains \(n\), so \[ \|x^A-x^B\|_\infty=1. \] Thus \[ \{x^A:A\subset\mathbb N\} \] is an uncountable subset of \(\ell^\infty\) whose distinct points are pairwise distance \(1\).
If \(\ell^\infty\) were separable, let \(D\) be a countable dense subset. The open balls \[ B(x^A,1/3) \] are pairwise disjoint, and density would require each such ball to contain a point of \(D\). This would inject the uncountable power set \(\mathcal P(\mathbb N)\) into the countable set \(D\), a contradiction. Therefore \(\ell^\infty\) is not separable.