Operator range category dichotomy

A bounded linear operator $T:X\to Y$ from a Banach space to a normed space either maps onto $Y$ or has meagre set image in $Y$. To prove this, write $T(X)={\bigcup }_{n\ge 1}T(n{\overline{B}}_X)$. If every $\overline{T(n{\overline{B}}_X)}$ has empty interior, each term is a nowhere dense set. Otherwise scaling and geometric correction for approximate surjectivity place a ball about zero inside $T({\overline{B}}_X)$; the vector subspace $T(X)$ then equals $Y$. In particular a nonclosed operator image is a meagre set. A proper closed operator image is also a meagre set, so the closed/meagre alternatives need not be exclusive.