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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

NazimJ
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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

NazimJ
  • 3,244