Are proper linear subspaces of Banach spaces always meager? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:46:37Z http://mathoverflow.net/feeds/question/3188 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager Are proper linear subspaces of Banach spaces always meager? Brandon Seward 2009-10-29T02:21:51Z 2011-04-17T15:25:52Z <p>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. </p> <p>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.</p> <p>More info:<br /> 1. Meager is just another word for first category, i.e. the countable union of nowhere dense sets.<br /> 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.<br /> 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.<br /> 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.</p> http://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager/3203#3203 Answer by Yemon Choi for Are proper linear subspaces of Banach spaces always meager? Yemon Choi 2009-10-29T03:29:16Z 2009-10-29T03:29:16Z <p>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.</p> http://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager/3614#3614 Answer by Jonas Meyer for Are proper linear subspaces of Banach spaces always meager? Jonas Meyer 2009-10-31T22:58:57Z 2009-10-31T22:58:57Z <p>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. </p> http://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager/3622#3622 Answer by Konstantin Slutsky for Are proper linear subspaces of Banach spaces always meager? Konstantin Slutsky 2009-10-31T23:48:48Z 2011-04-17T15:25:52Z <p>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 <a href="http://en.wikipedia.org/wiki/Discontinuous_linear_map" rel="nofollow">wikipedia</a>). 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.</p>