Clarification Regarding Banach space and Baire Category Theorem

Question

New to Baire Category. From a remark:

If we suppose that $Y$ is an infinite dimensional subspace of a Banach space $X$, and $Y$ has a countable (Hamel) basis, then one can show that $Y$ is first category, and also not closed.

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I'm a little unsure of why this follows. My thinking:

1. $Y$ is an infinite dimensional subspace, but not necessarily a Banach space. Hence, it's not necessarily complete.

2. Thus, it can have a countable Hamel basis (since Baire Category Theorem doesn't apply here).

3. So, $Y = \bigcup A_n$, is of first category since each $A_n$ is nowhere dense.

4. Any closed subspace of a Banach space is again a Banach space. Since $Y$ is not a Banach space, it cannot be closed.

Answer 1

If $Y$ has a countable Hamel basis $\{e_1,e_2,\ldots\}$, then let $A=\operatorname{span}\bigl(\{e_1,\ldots,e_n\}\bigr)$. Then $Y=\bigcup_{n\in\Bbb N}A_n$. But each $A_n$ is closed and $\mathring{A_n}=\emptyset$ . So, $Y$ is a countable union of closed subsets with empty interior, and so $Y$ is first category.

And, as you wrote, if it was closed, it would be complete.