# Borel ideals on $\omega$ are meager?

Let $\mathcal{I}$ be a proper ideal on $\omega$. If $\mathcal{I}$ is Borel as a subset of $2^\omega$, does it follow that $\mathcal{I}$ is meager?

Edit: What if $\mathcal{I}$ contains all finite subsets of $\omega$?

-
All sets not containing $0$ ... that's an ideal, right? And Borel. In fact clopen. So not meager. –  Gerald Edgar May 6 '13 at 13:06
Right. What if we assume that all finite subsets are in $\mathcal{I}$? –  jack May 6 '13 at 14:00

After googling, I found what "rational translations" means. But, Andreas how does that imply that $I$ and $F$ are comeager? Where did you use the fact that $I$ is Borel? –  jack May 6 '13 at 14:51
@jack: The assumption that I and therefore F are Borel is used to ensure that they have the Baire property, which in turn implies that, if they're non-meager, then they are comeager in some open sets. That plus closure under finite modifications makes them comeager in all of $2^\omega$. –  Andreas Blass May 6 '13 at 15:21
I see. Why is $F$ non-meager? –  jack May 6 '13 at 15:23