Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V$ be a complete normed space. Let $W$ be a proper vector subspace. Can $W$ be the intersection of a sequence of open dense subsets of $V$?

If there exists an open dense proper vector subspace then this problem would be silly. So as well as the problem above, I want to ask: why does there not exist an open dense proper vector subspace of a Banach space?

share|improve this question
    
Baire Category Theorem implies that such a $W$ have to be dense. If an open set $A$ contains a dense subspace, must $A=V$? –  Blythe Feb 21 '13 at 0:57

2 Answers 2

For your second question: An open proper subspace $Y$ of a normed space $X$ automatically satisfies $X = Y$ since it must contain a ball centered at zero, and thus any vector $x \in X$ (to see that: there must be a $\lambda > 0$ such that $\lambda x$ lies in that ball; now since $\lambda x \in Y$, we also have $x \in Y$.

share|improve this answer

No, $W$ can't be the intersection of open dense subsets by the following result of Pettis:

Let $G$ be a topological group and let $A \subseteq G$ be a non-meager subset having the Baire property. Then $A^{-1}A$ contains a neighborhood of the identity of $G$.

See Kechris, Classical descriptive set theory, Theorem (9.9), p.61 or Srivastava, A Course on Borel sets, Theorem 3.5.12, page 110f for proofs.

Assume that the subspace $W$ is dense in the Banach space $V$ and assume that $W$ is a countable intersection of open sets. By the Baire category theorem it is non-meager and $W$ has the Baire property. Since $W$ is a subspace, $W = W-W$, so $W$ contains a neighborhood of zero by Pettis's theorem. Since $W$ is invariant under translations by elements of $W$, it follows that $W$ is open, and since open subspaces are closed, $V = W$.

More generally, this argument shows that a dense subgroup with the Baire property of a connected topological group is either meager or the entire group. Nate Eldredge points out that assuming the Baire property is necessary here: The kernel of a discontinuous linear functional is dense, non-meager and has the Baire property, see this MO-post: http://mathoverflow.net/questions/3188/

share|improve this answer
1  
Nice answer, but your last sentence should be qualified: a dense subgroup with the Baire property is either meager or the entire group. For instance, this MO post shows that the kernel of a discontinuous linear functional on a Banach space is a linear subspace that is nonmeager and lacks the BP. –  Nate Eldredge Nov 6 '13 at 21:28
    
You're absolutely right. Thanks! –  Martin Nov 7 '13 at 6:19

Your Answer

 
discard

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.