Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

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 $k\in\mathbb{N} .$

share|cite|improve this answer
absolutely right. Thanks so much. – le duc quang Dec 10 '13 at 0:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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