# Can a proper vector subspace of a Banach space be a countable intersection of dense open subsets?

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?

-
 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 at 0:57

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 $x \in Y$.

-

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 of a connected topological group is either meager or the entire group.

-