If $T$ is an isomorphism acting on a separable Banach space, can we always find a countable dense subset $D$ of $X$ such that $T(D)=D? $
2026-04-09 07:17:41.1775719061
Every isomorphism on a separable Banach space has a completely invariant dense subset
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Summary of comments: yes, such $D$ can be obtained as $$D=\bigcup_{n\in\mathbb Z}T^n E$$ where $E$ is any countable dense subset of $X$.