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 $\omega$?
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 $\omega$? 


Let us consider a prime ideal J containing I. Suppose I is non meager. Then J would have non empty interior (modulo meager). Since complementing a set of integers is a meager preserving operation so the dual ultrafilter U of J has the same property. But both J and U are closed under rational translations and they both have interiors (modulo meager) so they are both comeager which is impossible. 

