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This seems like a fundamental result but I can not solve it of find an solution: Let $M:X\rightarrow U$ be a bounded linear map between Banach spaces. Show that if the range of M is a set of second category of U; then the range is all of U.

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    Should the title say "second category" instead of "second countable"?2012-11-29

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Assume that the range is of second category but the range of $M$ is not $U$. As $M(X)$ is not closed (otherwise $M(X)$ would be of first category, as it's a strict subspace), we can apply this result to $Y=\overline{M(X)}$.

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    I don't know, but what we have to show is strongly related to what is done in the link.2012-11-30