Excercise 15 Rudin functional analysis chapter 2

Question

I am self-studying the book function analysis of Rudin. I got stuck on the final passage of the following exercise.

Suppose $X$ is an $F-$space (a topological vector space with a topology induced by a complete metric) and $Y$ is a subspace of $X$ whose complement is of the first category. Prove that $Y=X.$ Hint: $Y$ must intersect $x+Y$ for every $x$ in $X.$

I understand that the hint it is just a convenient way to restate: $x \in Y$ for every $x \in X.$ The first thing I thought is to use Baire category theorem. First, I exploit that $Y$ has a complement, let us call it $C$, then we can write $$ X=\bigcup_{c \in C}c+Y.$$ My questions are: Is $C$ countable?; Is there a countable subset $B$ of $C$ for which we still have
$$ X=\bigcup_{c \in B}c+Y?$$ If so, then I would have finished because by the Baire category theorem there has to be some set $c+Y$ with non empty internal part, and an affine subspace with non empty internal part has to coincide with the whole space.

Answer 1

Is $C$ countable?

No, it's often not. Assuming $C$ is non-trivial and the ground field is uncountable (e.g. $\Bbb{R}$), then $C$ is definitely not countable.

Is there a countable subset $B$ of $C$ for which we still have $$X=\bigcup_{c∈B} c+Y?$$

No, for the most part. Indeed, one cannot throw away any elements from $C$ and still get $X$ from this union (note that every $x \in X$ is a unique sum of elements from $Y$ and from $C$).

Instead, consider this: if $Y$ and $x + Y$ had empty intersection, then their respective complements, $X \setminus Y$ and $X \setminus (x + Y)$ would union to produce the entire space. We know that $X \setminus Y$ is already a countable union of nowhere dense sets. If you can show the same for $X \setminus (x + Y)$, then this would make $X$ of the first category, contradicting the Baire Category Theorem. See if you can finish for yourself.