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.

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

4 Answers 4


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 $X$.

  • $\begingroup$ This is very interesting. Thanks for taking the time to post, and pointing out the error that at least 7 of us missed! $\endgroup$ Jul 8, 2014 at 16:50
  • $\begingroup$ 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?. $\endgroup$ 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.

  • $\begingroup$ 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 $y = (y-az) + az \in B_i$. $\endgroup$ 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.

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

  • $\begingroup$ 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. $\endgroup$ Oct 29, 2009 at 3:36
  • $\begingroup$ thanks, and you're right about the rationals, I misread your question. Sorry! $\endgroup$
    – Yemon Choi
    Oct 29, 2009 at 3:56

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