1
$\begingroup$

In reading Kechris book.

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.

$\endgroup$
2
$\begingroup$

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 $\sim \bigcap_{n<\omega}\sim A_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.