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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 $\omega$?

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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
up vote 2 down vote accepted

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.

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Much as I like ultrafilters, I have to point out that they're not really needed here. If I is a proper, non-meager, Borel ideal that contains the finite sets, and if F is the dual filter, then, being closed under finite modifications (a.k.a. rational translations), both I and F are comeager, which is absurd because they are disjoint. – Andreas Blass May 6 '13 at 14:27
Hi. I'm new to descriptive set theory, so could you use layman's language? I don't understand the terms "prime ideal", "modulo meager", and "rational translations". – jack May 6 '13 at 14:34
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

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