Definition. Let $X$ be a topological space and $U\subset X$ an open set. We say that $A$ is meager in $U$ if $A\cap U$ is meager in $X.$ Also, $A$ is co-meager in $U$ if $U\setminus A$ is meager.
Let $\Bbb B$ be a countable basis of $\Bbb R.$ Let $\{P^I\subset I\colon I\in\Bbb B\}$ be pairwise disjoint meager set of $\Bbb R$. Then $$I\setminus\bigcup_{I\in\Bbb B} P^I$$ is co-meager in $I.$ Indeed, it is enough to show that $\bigcup_{I\in\Bbb B} P^I= I\setminus\bigg(I\setminus\bigcup_{I\in\Bbb B} P^I\bigg)$ is meager in $I.$ To see this we need to check that $\bigcup_{I\in\Bbb B} P^I\cap I$ is meager in $\Bbb R.$ Of course, this is true since $P^I=\bigcup_{I\in\Bbb B} P^I\cap I$ is meager in $\Bbb R$ by the assumption.
Is that true? I just need to make sure that I used the definition in the correct way.
Thank you in advance.
Best Answer
$I$ is Polish and so a countable union of meagre sets in it is still meagre. That's the whole proof.