# Question about of Baire property and Baire space

Please, I would like help with this proposition.

For convencion we put for $A \subseteq X$,

$$\sim A=X\setminus A$$

If $A$ is comeager in $U$, we say that $U$ forces $A$, in symbols $$U \Vdash A$$

A weak basis for a topological space $X$ is a collection of nonempty open sets such that every nonempty open set contains one of them

Proposition Let $X$ be a topological space.

$(i)$ If $X$ is a Baire space, $A$ has the Baire property in $X$ and $U \subseteq X$ open set and $V$ over a weak basis, then

$$U \Vdash \sim A \Leftrightarrow \forall{V}\subseteq U (V \nVdash A)$$

$(ii)$ If $X$ is a Baire space, $A_n \subseteq X$ have the Baire property, and $U \subseteq X$ open sets and $V,W$ over weak basis, then

$$U \Vdash \bigcup_n A_n \Leftrightarrow \forall{V} \subseteq U \exists{W} \subseteq V \exists {n} \Vdash A_n.$$

Proof: $(i)$ Note that if $U \subseteq X$ is open, then $A \cap U$ has the Baire property in $U$. Why $(ii)$ follows from this fact?

$(ii)$ Any ideas.

Proposition Let $X$ be a topological space and suppose that $A\subseteq X$ has the Baire property. Then either $A$ is meager or there is a nonempty open set $U\subseteq X$ on which $A$ is co-meager. If $X$ is a Baire space, exactly one these alternatives holds.(This is the proposition $8.26$ Book Kechris).
For $(i)$ You must apply the above proposition to $U$ instead of $X$ (why you need that $A\cap U$ has the BP in $U$). As $X$ is Baire, $U$ also is a Baire space, so that the dichotomy of the proposition is true (i.e, that both of disjunction are not met at the time). Remember to apply the weak base.
For $(ii)$ can be done using $(ii)$, since $\bigcup_{n<\omega}A_n$ equals $\sim \bigcap_{n<\omega}\sim A_n$.