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It is said here that since an infinite-dimensional Banach space $M$ is meagre (it is contained in the countable union of nowhere dense closed subsets of itself), we reach a contradiction. However, I did not exactly understand what the contradiction is. Is it that the interior of a Banach space is necessarily non-empty, since it is open?

Just want to make sure I get it right.

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    $\begingroup$ Baire category theorem states that any complete metric space is of second category, that is, comeagre. $\endgroup$ – Idonknow Nov 29 '17 at 10:36
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Every finite dimensional subspace is closed (why?). Every proper subspace has empty interior (why?). So in an infinite dimensional space every finite dimensional subspace is nowhere dense. A countable dimensional subspace is a union of countably many finite dimensional subspaces (take span of the first $k$ basis vectors for each $k$—a nested union). So every countably-infinite dimensional subspace is first category. But the whole Banach space is second category by Baire’s theorem.

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  • $\begingroup$ I think that the reason every finite-dimensional subspace is closed is because it is spanned by a finite set of vectors. One can also show that in a finite-dimensional subspace every absolutely summable series is summable, so the space is complete and hence closed. Every proper subspace of a normed vector space has empty interior because a proper subspace if it is closed, so it has no epsilon-balls contained in it. $\endgroup$ – sequence Nov 29 '17 at 11:21

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