# Baire Category Theorem in infinite- dimensional space

Let $$X$$ be an infinite-dimensional Banach space. Suppose that $$M$$ is an infinite- dimensional subspace of $$X$$ that has a countable Hamel basis. I want to show that $$M$$ is a meager subset of $$X$$.

My attempt: Since $$M$$ has countable Hamel basis, let $$\{x_n\}_{n \in \mathbb{N}}$$ be a countable Hamel basis of $$M$$ and let $$F_n=\text{span}\{x_1,...,x_n\}$$. Then $$F_n$$ is nowhere dense. Then I want to use Baire Category Theorem however it requires the completeness of $$M$$. But if $$M$$ is complete then it's a infinite-dimensional Banach space, and we know there won't be countable Hamel basis in $$M$$. Which part was I wrong? Any help or hint is appreicated.

• You have $M=\cup_i F_i$ and that's enough to conclude. – Shivering Soldier Apr 26 at 14:55
• Doesn't Baire Category Theorem only applicable in nonempty complete metric spaces? – Maskoff Apr 26 at 14:58
• A Banach space is complete. You need the completeness of $X$. – Henno Brandsma Apr 26 at 15:01
• @HennoBrandsma But I was trying to use $M=\bigcup_{n=1}^{\infty} F_{n}$, shouldn't the comleteness of $M$ required? – Maskoff Apr 26 at 15:04

Well, $$M$$ is by definition a countable union of nowhere dense sets, so meagre. This is without completeness, just by definitions. There is a bit of work to see that the $$F_n$$ are closed and have empty interior (as $$X$$ is infinite-dimensional), but that should be standard.
Completeness of $$X$$ then tells us that $$M \neq X$$, even has empty interior. So $$X$$ cannot have a countable Hamel base, which is usually the point.