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.

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

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