Theory of (definable) ideals on a multi-dimensional countable set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:20:15Z http://mathoverflow.net/feeds/question/107354 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107354/theory-of-definable-ideals-on-a-multi-dimensional-countable-set Theory of (definable) ideals on a multi-dimensional countable set Justin Palumbo 2012-09-17T00:55:01Z 2012-09-17T00:55:01Z <p>I'm wondering if there are any results about definable ideals/filters on $\omega\times\omega$, $\omega\times\omega\times\omega$, etc.</p> <p>To give a sense of the kind of results I might be looking for: one shining example of a result on definable ideals on a countable set is the following theorem due to Talagrand: (note that definable ideals must have the Baire property and hence b/c of 0-1 laws must be meager)</p> <p>Theorem. Suppose $\mathcal{I}$ is an ideal on $\omega$ which is meager when viewed as a subset of $2^\omega$. There is a partition $\omega=\bigcup_{n&lt;\omega}I_n$ into intervals so that any set containing infinitely many intervals does not belong to $\mathcal{I}$.</p> <p>A natural two-dimensional analogue of this theorem would say that given a definable ideal $\mathcal{I}$ on $\omega\times\omega$ there is a pair of interval partitions $\omega=\bigcup_{n&lt;\omega}I_n$ and $\omega=\bigcup_{n&lt;\omega}J_n$ such that whenever $A\subseteq\omega\times\omega$ and ($\exists^\infty m)(\exists^\infty n) I_m\times J_n\subseteq A$ then $A$ does not belong to $\mathcal{I}$. This two-dimensional analogue does not hold generally for definable ideals but it seems conceivable there are some fairly general situations where it does.</p> <p>But this is just an example; I'd be interested in any result that explores the properties definable ideals on $\omega^k$ have as multidimensional objects.</p> <p>I haven't been able to find anything along these lines in the literature, but it's very possible I've been looking in the wrong places.</p>