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We have $X$ an $F$-space, $Y$ a subspace of $X$ whose complement is of the first category. Prove that $Y=X$. This problem have a hint: need to show that $Y $ intersect $x+Y$ $\forall x \in X$, but i still don't know how do it. Thanks in advace.

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The first sentence led me to think you intended the topological meaning of "F-space", but then the mention of addition in the hint leads me to think that perhaps $F$ is supposed to be a field, $X$ a vector space over it, and $Y$ a vector subspace. – Andreas Blass Jan 8 '13 at 19:08
@Andreas: I suspect that this notion of $F$-space may be intended. – Brian M. Scott Jan 8 '13 at 19:09
@AndreasBlass: I agree with Brian M.Scott. Sorry about that. – user52523 Jan 8 '13 at 19:13
Can you show that $x+Y$ is co-meagre for each $x\in X$? If you can do that, you can use the fact that the union of two meagre sets is meagre to show that $Y\cap(x+Y)\ne\varnothing$. – Brian M. Scott Jan 8 '13 at 19:14
You need the fact that an $F$-space is a complete metric space in order to use the Baire category theorem, and you can use translation-invariance of the metric to show that $x+Y$ is co-meagre. – Brian M. Scott Jan 8 '13 at 19:58
up vote 1 down vote accepted

It is clear that for every $x \in X$, $(x+Y) \cap Y \neq \emptyset$. Indeed, let us suppose that there exists $x \in X$ s.t. $x+Y \subset X \setminus Y$. Then $x+Y$ is of the first category, hence - since addition is a linear homeomorphism - $Y$ is of the first category. This is absurd, because we would have $X=Y \cup (X \setminus Y)$ so $X$ of the first category, absurd by Baire ($X$ is a complete metric space).

Now we have that for every $x \in X$ there exists $y \in Y \cap (x+Y)$, i.e. there exists $y'$ s.t. $y=x+y'$, hence $x=y-y' \in Y$ because $Y$ is a linear subspace. Then $X \subseteq Y$, which implies $X=Y$.

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