No infinite-dimensional $F$-space has a countable Hamel basis

Question

If $X$ is an infinite-dimensional topological vector space which is the union of countably many finite-dimensional subspaces, prove that $X$ is of the first category in itself. Prove that therefore no infinitely-dimensional F-space has a countable Hamel basis

This is an exercise from Rudin book. But I can't construct a set of nowhere dense closed sets which have union equal $X$. Can anyone help me? Thanks.

Hint: The linear span of the first $n$ vectors is nowhere dense. — Chris Janjigian, Dec 09 '13 at 15:15

score 9 · Accepted Answer · answered Dec 09 '13 at 15:19

9

Let $\{e_n :n\in\mathbb{N}\}$ is countable basis of $F$ and let denote by $$F_n =\{\lambda_1 e_1 +\lambda_2 e_2 +...+\lambda_n e_n :\lambda_1 ,\lambda_2 ,...,\lambda_n \in\mathbb{R}\} .$$ Then $$F=\bigcup_{j=1}^{\infty} F_j .$$ Since $F_n$ are closed and $F$ is complete then by Baire theorem we have that $F\subset F_k $ for some $k\in\mathbb{N} .$

answered Dec 09 '13 at 15:19

absolutely right. Thanks so much. – le duc quang Dec 10 '13 at 00:57
Which Baire theorem are you using ?@leducquang – Devendra Singh Rana Apr 24 '19 at 05:58

No infinite-dimensional $F$-space has a countable Hamel basis

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