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.

  • $\begingroup$ Hint: The linear span of the first $n$ vectors is nowhere dense. $\endgroup$ – Chris Janjigian Dec 9 '13 at 15:15
  • $\begingroup$ Does $F$-space mean a Fréchet space? $\endgroup$ – tomasz Dec 9 '13 at 15:15
  • $\begingroup$ @tomasz: F-space is a complete metric space. $\endgroup$ – le duc quang Dec 10 '13 at 0:58
  • $\begingroup$ @leducquang: then what do you mean by infinite-dimensional? $\endgroup$ – tomasz Dec 10 '13 at 3:24
  • $\begingroup$ Sorry, I mean the complete metric topological space. $\endgroup$ – le duc quang Dec 10 '13 at 12:07

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 $k\in\mathbb{N} .$

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  • $\begingroup$ absolutely right. Thanks so much. $\endgroup$ – le duc quang Dec 10 '13 at 0:57
  • $\begingroup$ Which Baire theorem are you using ?@leducquang $\endgroup$ – Devendra Singh Rana Apr 24 '19 at 5:58

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