= 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 $T(X)=\bigcup_{n\geq1}T(n\overline B_X)$. 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.
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