# Are proper linear subspaces of Banach spaces always meager?

Let X be a Banach space, and let Y be a proper non-meager linear subspace of X. If Y is not dense in X, then it is easy to see that the closure of Y has empty interior, contradicting Y being non-meager. So Y must be dense. If Y has the Baire property, then it follows from Pettis Lemma that Y is open and hence closed (since the complement of Y is the union of translates of Y), contradicting Y being proper. Thus, Y must be dense and not have the Baire property.

My question is: is there a Banach space X with a proper non-meager linear subspace Y? Such a Y must be dense and not have the Baire property. Any such Y must be difficult to construct since all Borel sets and even all continuous images of separable complete metric spaces have the Baire property.

1. Meager is just another word for first category, i.e. the countable union of nowhere dense sets.
2. A set A in a topological space has the Baire property if for some open set V (possibly empty) the set (A-V)U(V-A) is meager.
3. The collection of sets with the Baire property form a sigma-algebra. All open sets trivially have the Baire property, thus all Borel sets have the Baire property. All analytic sets also have the Baire property.
4. Pettis Lemma: Let G be a topological group and let A be a non-meager subset of G with the Baire property. Then the set A*A^{-1} (element-wise multiplication) contains an open neighborhood of the identity. This is an analog to a similar theorem about Lebesgue measure: If A is a Lebesgue measurable subset of the reals with positive Lebesgue measure, then A - A (element-wise subtraction) contains an open set around 0.

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In what topology? –  Qiaochu Yuan Oct 29 '09 at 3:41

I think you can have such a subspace. Let $f : X \to R$ be a discontinous linear functional (such a functional exists assuming Axiom of Choice, see wikipedia). The claim is that its kernel $K = \ker f$ is a proper non-meager subspace. It is definitely proper. Assume it would be meager. Then it is contained in the countable union of closed subsets $A_i$. Since $K=\ker f$ it has codimension 1, so there is $z \in X$ such that every $x \in X$ can be written as $x = k + az$, for some number $a$ and some $k \in K$. Let $B_i$ be the set of elements $A_i + [-i,i]z$ (that is the set of $x \in X$ for which $x = k+az$ where $k$ is in $A_i$ and $a$ in $[-i,i]$). Then, I think, $B_i$ are closed and they have to have empty interior. Indeed, if there is a small ball around $k + az$ in $B_i$ then $f$ will be continuous at $k+az$, and then (since $f$ is linear) continuous everywhere, contradicting the choice of $f$. Thus $B_i$ are closed and nowhere dense, but their union is then the whole space $X$, which contradicts Baire's theorem.

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The $B_i$ are indeed closed. Suppose $x_n + a_n z \to y$ where $x_n \in A_i$ and $a_n \in [-i,i]$. Since $[-i,i]$ is compact, by passing to a subsequence we may assume $a_n \to a$. Thus $x_n \to y - az \in A_i$ since $A_i$ is closed. So $y = (y-az) + az \in B_i$. –  Nate Eldredge Apr 17 '11 at 15:28

This is more of a question than an answer, but hopefully it helps. What happens if Y is the kernel of a discontinuous linear functional on X? Such functionals are easy to "construct" using Zorn's Lemma for the existence of a linear basis for X (X infinite dimensional of course). In that case, Y is not closed and has codimension 1, so it is dense. It seems to me that it would be non-meager, but I don't have an argument for why it is non-meager.

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I am sorry, I didn't read you comment before posting mine. My fault. –  Konstantin Slutsky Oct 31 '09 at 23:57
Don't apologize; my answer wasn't complete, and you answered my question. Thanks. –  Jonas Meyer Nov 1 '09 at 0:09