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In Kechris's book "Classical Descriputive set theory" Chapter 23 (Exercise 23.4), it claim that there is a $\Sigma^0_\xi$-complete ideal on $\omega$ for each $\xi\geq 3$.

There is a candidate construction as follows: give a $\Sigma^0_\xi$-complete $A$ subset of $2^\omega$, define an ideal $\mathcal I_A$ on ${}^{<\omega}2$ generated by $\{\{x|n:n\in\omega\}:x\in A\}$. I don't underatand why the Borel hierarchy of $\mathcal I_A$ isn't bigger than $\xi$?

Is there a different construction?

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  • $\begingroup$ I think you should try to show that the maps $(\{x_1, x_2, \dots, x_n\}, y_1, y_2, \dots, y_n) \mapsto z_1 \cup z_2 \cup \dots \cup z_n$ where $z_i = x_i \upharpoonright y_i$ are sufficiently injective on $[P]^n \times (2^{\omega})^n$ ($P$ is a perfect almost disjoint family) in order to preserve sufficiently high Borel ranks. $\endgroup$
    – Ashutosh
    Commented May 31, 2015 at 23:02
  • $\begingroup$ I can find a boud rank for $\mathcal I_A$ as follows: Define $X \subseteq \mathcal P({}^{<\omega}2)$ with $x\in X$ iff $\exists n \forall a\in [{}^{<\omega}2]^n((\forall s\in a\Rightarrow s\in x)\Rightarrow (\exists s\not= t\in a (s\prec t \vee t\prec s )))$. Define $g:X\longrightarrow [2^\omega]^{<\omega}$ with $g(x)=\{y\in 2^\omega:\forall n\exists m> n y|m\in x\}$, then $\mathcal I_A=g^{-1}([A]^{<\omega})$, but I can't compute the rank of $\mathcal I_A$ is $\xi$. $\endgroup$
    – Jialiang He
    Commented Jun 1, 2015 at 1:45
  • $\begingroup$ And the other directed is easy, the map $f:2^\omega \longrightarrow \mathcal P({}^{<\omega}2)$ with $f(x)=\{\{x|n\}:n\in\omega\}$, then $ A=f^{-1}(\mathcal I_A)$. $\endgroup$
    – Jialiang He
    Commented Jun 1, 2015 at 1:50

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