Szymon Żeberski: Mycielski theorem and Miller trees
04/04/19 15:37
Tuesday, April 9, 2019 17:15
Room: D1-215
Speaker: Szymon Żeberski
Title: Mycielski theorem and Miller trees
Abstract. The classical Mycielski theorem says that for comeager \(A\subseteq [0,1]^2\) one can find a perfect set \(P\) such that \(P\times P\subseteq A\cup\Delta\). (The same is true if we start with \(A\) of measure 1.)
We will discuss how far this can be generalized if we replace perfect set by superperfect set, i.e a body of a Miller tree.
It turns out that there is a comeager \(A\subseteq (\omega^\omega)^2\) such that \(A\cup \Delta\) does not contain any set of the form \(M\times M\), where \(M\) is superperfect.
However, for comeager \(A\subseteq [0,1]^2\) one can find a perfect set \(P\) and a superperfect set \(M\supseteq P\) such that \(P\times M\subseteq A\cup\Delta\).
We will also discuss measure case, where results are slightly different.
Room: D1-215
Speaker: Szymon Żeberski
Title: Mycielski theorem and Miller trees
Abstract. The classical Mycielski theorem says that for comeager \(A\subseteq [0,1]^2\) one can find a perfect set \(P\) such that \(P\times P\subseteq A\cup\Delta\). (The same is true if we start with \(A\) of measure 1.)
We will discuss how far this can be generalized if we replace perfect set by superperfect set, i.e a body of a Miller tree.
It turns out that there is a comeager \(A\subseteq (\omega^\omega)^2\) such that \(A\cup \Delta\) does not contain any set of the form \(M\times M\), where \(M\) is superperfect.
However, for comeager \(A\subseteq [0,1]^2\) one can find a perfect set \(P\) and a superperfect set \(M\supseteq P\) such that \(P\times M\subseteq A\cup\Delta\).
We will also discuss measure case, where results are slightly different.