Closed convex absorbing set has an origin neighbourhood

Let $S$ be a nonempty closed, convex, symmetric subset of a Banach space and suppose ${\bigcup }_{n\ge 1}nS=X$. Baire's theorem puts a ball $B(x,r)$ inside some $nS$. Symmetry also puts $B(-x,r)$ there, and convexity puts the midpoint of $x+z$ and $-x+z$, namely $z$, in $nS$ whenever $\Vert z\Vert <r$. Hence $B(0,r/n)\subset S$.

Convexity is essential. In $\mathbb{R}$, letThis set is closed, symmetric, and has gaps arbitrarily near zero. Given $x\ne 0$, choose $k$ so small-scale that the interval
$[|x|/(2\cdot 4^{-k}),|x|/4^{-k}]$ has length at least one; it contains an integer $n$, and then $x/n\in S$. Thus ${\bigcup }_{n}nS=\mathbb{R}$, although $S$ is not a neighbourhood of zero.