Timeline for $\Delta^0_{\alpha}$ universal sets does not exist

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Oct 9, 2011 at 22:36	comment	added	Asaf Karagila♦		Closely reading your answer it seems that you prove that $\Sigma^0_\alpha\neq\Pi^0_\alpha$ and for $\beta<\alpha$ we have $\Sigma^0_\alpha\subsetneq\Delta^0_\beta$. This is not what I asked about. My question is why for $\alpha>1$ there is no $\Delta^0_\alpha$ subset $A\subseteq X\times X$ such that for every $x\in X$ the cut $\lbrace y\mid (x,y)\in A\rbrace$ is a $\Delta^0_\alpha$ set, and for every $\Delta^0_\alpha$ set $Y\subseteq X$ there is some $x\in X$ such that $\lbrace x\rbrace\times Y\subseteq A$.
Oct 9, 2011 at 21:16	history	edited	Rachid Atmai	CC BY-SA 3.0	added 7 characters in body
Oct 9, 2011 at 20:27	comment	added	Asaf Karagila♦		You meant $\alpha<\omega_1$ in that last part of the theorem statement? Since $\alpha<1\Rightarrow\alpha=0$... :-)
Oct 9, 2011 at 19:27	history	answered	Rachid Atmai	CC BY-SA 3.0