# 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.

• In what topology? Oct 29, 2009 at 3:41
• What about this: Take any Baire normed space Y which is not complete and let X be the completion of Y.
– user171241
Dec 26, 2020 at 11:24

I am afraid that Konstantin's accepted answer is seriously flawed.

In fact, what seems to be proved in his answer is that $\ker f$ is of second category, whenever $f$ is a discontinuous linear functional on a Banach space $X$. This assertion has been known as Wilansky-Klee conjecture and has been disproved by Arias de Reyna under Martin's axiom (MA). He has proved that, under (MA), in any separable Banach space there exists a discontinuous linear functional $f$ such that $\ker f$ is of first category. There have been some subsequent generalizations, see Kakol et al.

So, where is the gap in the above proof?

It is implicitly assumed that $\ker f = \bigcup A_i$. Then $f$ is bounded on $B_i=A_i+[-i,i]z$. But in reality, we have only $\ker f \subset \bigcup A_i$ and we cannot conclude that $f$ is bounded on $B_i$.

And finally, what is the answer to the OP's question?

It should not be surprising (remember the conjecture of Klee and Wilansky) that the answer is: in every infinite dimensional Banach space $X$ there exists a discontinuous linear form $f$ such that $\ker f$ is of second category.

Indeed, let $(e_\gamma)_{\gamma \in \Gamma}$ be a normalized Hamel basis of $X$. Let us split $\Gamma$ into countably many pairwise disjoint sets $\Gamma =\bigcup_{n=1}^\infty \Gamma_n$, each of them infinite. We put $X_n=span\{e_\gamma: \gamma \in \bigcup_{i=1}^n \Gamma_i\}$. It is clear (from the definition of Hamel basis) that $X=\bigcup X_n$. Therefore there exists $n$ such that $X_n$ is of second category. Finally, we define $f(e_\gamma)=0$ for every $\gamma \in \bigcup_{i=1}^n\Gamma_i$ and $f(e_{\gamma_k})=k$ for some sequence $(\gamma_k) \subset \Gamma_{n+1}$. We extend $f$ to be a linear functional on $X$. It is clearly unbounded, $f\neq 0$, and $X_n \subset \ker f$. Hence $\ker f\neq X$ is dense in $X$ and of second category in $X$.

• This is very interesting. Thanks for taking the time to post, and pointing out the error that at least 7 of us missed! Jul 8, 2014 at 16:50
• For a survey of related issues (Borel types of linear subspaces in infinite dimensional Banach spaces), see my answer to the math StackExchange question Does there exist a linearly independent and dense subset?. Jul 8, 2014 at 17:50

## EDIT: The following argument is in error. See Tony Prochazka's answer.

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.

• 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 $y = (y-az) + az \in B_i$. Apr 17, 2011 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.

• I am sorry, I didn't read you comment before posting mine. My fault. Oct 31, 2009 at 23:57
• Don't apologize; my answer wasn't complete, and you answered my question. Thanks. Nov 1, 2009 at 0:09

By subspace, do you mean "linear subspace"? Because if not, I don't see what's wrong with taking X to be the real line and Y to be the rationals.

• Yes, that's right, linear subspace. I changed it so no one else will be confused. But even without requiring linearity, the rationals don't work. You can express the rationals as a countable union of one point sets. Oct 29, 2009 at 3:36
• thanks, and you're right about the rationals, I misread your question. Sorry! Oct 29, 2009 at 3:56