So I got $0$ credit on one of my school problems. The comment was that if $Y$ is infinite dimensional, I cannot assume it has a countable basis. Hence I have absolutely no ideas how to solve this problem. Please help so I can learn how to handle this type of scenario..
Given $X$ a completely metrizable TVS, with proper linear subspace $Y$. Can $X \setminus Y$ be of first category? (Recall a set is of first category if it is the countable union of nowhere dense sets)
I will include my answer below, just to show my thought process...
$X$ is completely metrizable, so by Baire Category Theorem it cannot be the countable union of nowhere dense sets. I will show that every proper linear subspace $Y$ is the countable union of nowhere dense sets, meaning $X\setminus Y$ cannot be first categroy. Otherwise $X=Y\cup (X \setminus Y)$ is the union of countably many nowhere dense sets, a contradiction
Assume that $Y\subseteq X$ is a finite dimensional, proper, linear subspace. Then I show that it is closed with empty interior, hence nowhere dense $(*)$
Then assume that $Y\subseteq X$ is an infinite dimensional, proper, linear subspace. Since it is linear, it has a basis $Y = span\{ y_1, y_2, ... \}$. Define the following sets for each $n\in\mathbb{N}$: $$ S_n=span\{ y_1 , ..., y_n \} $$ Each $S_n$ is nowhere dense by $(*)$, and $$ Y = \bigcup_{n=1}^{\infty}S_n $$ Hence $Y$ is first category
The closest I've got is this, from the link in the comment of Dave L. Renfro. Not sure yet how applicable it is, still have to think about it more
$X$ is completely metrizable, so by Baire Category Theorem it cannot be the countable union of nowhere dense sets. I will show that every proper linear subspace $Y$ is nowhere dense sets, meaning $X\setminus Y$ cannot be first categroy. Otherwise $X=Y\cup (X \setminus Y)$ is the union of countably many nowhere dense sets, a contradiction
Given $Y$ a proper linear subspace of $X$, assume by contradiction that $Y$ is not nowhere dense. Then it contains a non-empty open set $V$. Since $X$ is a TVS and continuous under addition, we can find a balanced open set $U$ contained in $V$ . So far we have $$ U \subseteq V \subseteq Y$$ Since $Y$ is linear, $tU \subseteq Y$ for all $t\in\mathbb{R}$. This is a contradiction to the fact that $Y$ is a proper subset of $X$. So $Y$ is nowhere dense