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.

  • 1
    $\begingroup$ Baire category theorem states that any complete metric space is of second category, that is, comeagre. $\endgroup$ – Idonknow Nov 29 '17 at 10:36

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.