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I have started reading about Borel Hierarchies. If I understand correctly, Then $\Sigma_2^0$ is the collection of all sets of the form $A=\bigcup_{n \in \omega}{B_n}^c$, where $B_n$ is open. also, $\Sigma_3^0$ is the collection of all sets of the form $A=\bigcup_{n \in \omega}{B_n}^c$, where $B_n$ is in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$. I can think of a set which is in $\Sigma_{3}^0$ and not in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$, for example, the set, $(-\infty,0] \cup \bigcup_{n \in \mathbb{N}} (\frac{1}{n+1},\frac{1}{n}) \cup (1,\infty)$. But, is there an example of a set which is in $\Sigma_{4}^0$ but not in $\Sigma_{3}^0$ or in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$?

Thank you! Shir

Glorfindel
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topsi
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1 Answers1

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You cannot construct a $\Sigma_4^0$ Borel set, for one major reason, it is consistent with the failure of choice that every set is $\Sigma_4^0$.

The main method of showing that there are such sets is to prove the existence of $\Sigma^0_n$-universal sets. You can read my question, and answer, about those here.

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Asaf Karagila
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  • This is getting more and more fascinating.. I am not sure I yet have the mathematical maturity to fully understand what you wrote there but I'll give it a try. Thank you! – topsi Dec 23 '13 at 11:46
  • Of course you can construct the sets, Asaf. What you cannot do is to prove without some appeal to choice that they are properly $\Sigma^0_4$. – Andrés E. Caicedo Dec 23 '13 at 15:49
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    @Andres: Potato, tomato. Both are from the solanum family. :-) I suppose it's that time of the year where I'm somewhat more absent minded and less careful in my choice of words. I'll try to pay more attention to that in the future. – Asaf Karagila Dec 23 '13 at 15:56
  • You mean "it is consistent with the failure of choice that every set is $\Sigma^0_3$", (not $\Sigma^0_4$), right? – Alex Kruckman Jan 23 '19 at 00:55
  • @Alex: The question was about $\Sigma^0_4$. I am not claiming this is optimal. – Asaf Karagila Jan 23 '19 at 07:07
  • @Asaf Sure! I just meant that to show that you need choice to exhibit a properly $\Sigma^0_4$ set, you need to know that it's consistent with ZF that every $\Sigma^0_4$ set is $\Sigma^0_3$. – Alex Kruckman Jan 23 '19 at 14:18