Separable Banach space

Remarks

Equivalently, there is a countable set $S \subseteq X$ whose linear span is dense in $X$. Indeed, finite combinations of elements of $S$ with rational (or Gaussian rational) coefficients then form a countable dense set. Example: $\ell^1$ is separable. The finitely supported sequences with rational entries form a countable set $D$. For $x \in \ell^1$ and $\varepsilon > 0$, choose $N$ with $\sum_{n > N} \abs{x_n} < \varepsilon/2$. Then choose rational $q_n$ with $\sum_{n \le N} \abs{x_n - q_n} < \varepsilon/2$. Then $(q_1, \ldots, q_N, 0, \ldots) \in D$ is within $\varepsilon$ of $x$. Non-example: $\ell^\infty$ is not separable. For each subset $A$ of $\N$ its indicator sequence $\mathbf{1}_A$ lies in $\ell^\infty$, and $\norm{\mathbf{1}_A - \mathbf{1}_B}_\infty = 1$ for $A \ne B$. The open balls of radius $1/2$ around these uncountably many points are disjoint, so no countable set meets them all. If the dual $X^*$ is separable, then $X$ is separable; the converse fails, as $X = \ell^1$ with $X^* = \ell^\infty$ shows.