0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.