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I am taking a course in descriptive set theory, and the exam is approaching on Sunday. In the framework of proving that for an uncountable Polish space $X$ the following holds: $\Delta^0_\alpha(X)\subsetneqq\Sigma^0_\alpha(X),\Pi^0_\alpha(X)\subsetneqq\Delta^0_\beta(X)$.

We did not only prove that there is a set which is $\Sigma^0_\alpha\setminus\Pi^0_\alpha$ but also proved that for every uncountable Polish space there are no $\Delta^0_\alpha$-universal sets based on any Polish space.

The proof is very simple that for $X$ there is no universal set based on $X$ itself, as well for $\alpha>2$ the proof is quite simple. For $\alpha=2$ the proof given to us goes through great lengths in a rather complex proof.

Is there a rather general and relatively simple statement which does not separate the cases $\alpha=2$ and $\alpha>2$?

(I could not find the general theorem in either Kechris nor Moschovakis, but only for the case we wish to base the universal set on $X$)

Edit: I am somewhat under the impression that this theorem has not been published before. My teacher claims that it is unlikely that a unified argument will hold for $\alpha\ge 2$ and separation must be made. I'm still not 100% convinced.

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1 Answer 1

Hi. Here's the proof I've learned. It does not separate the cases $\alpha=2$ and $\alpha>2$. I hope I am not misunderstanding your question.

Theorem: Let $X$ be an uncountable Polish space. Then for any $\alpha<\omega_1$, there is a set $A\in \Sigma^0_{\alpha} \backslash \Pi^0_{\alpha}$. There is also for each $\alpha<\omega_1$ a set in $\Delta^0_{\alpha} \backslash \bigcup_{\beta<\alpha} \Sigma^0_{\beta} $

Proof: Let $X$ be an uncountable Polish space. The theorem is quick in the case $\alpha=1$ since if every open set was also closed then every point and every subset of $X$ is open which gives $X$ the discrete topology. But $X$ was Polish. Let $\alpha>1$. Recall that is $X$ is uncountable Polish then there is an embedding of $2^{\omega}$ into $X$. So there is a subspace $C \subseteq X$ homeomorphic to the Cantor space. $C$ has to be closed in $X$. Since having a universal set implies non-self-duality and by the existence of universal sets, there is a set $A\subseteq C$ which is $ \Sigma^0_{\alpha} $ but not $ \Pi^0_{\alpha} $ in the subspace topology on $C$. So in $X$, $A$ is the intersection of $ \Sigma^0_{\alpha} $ set and a closed set so it is $ \Sigma^0_{\alpha} $. If $A$ were also $\Pi^0_{\alpha} $ then $A=A \cap C$ would also be $ \Pi^0_{\alpha} $ in the subspace topology on $C$ which is a contradiction. For the second part, let $2\leq \alpha$. We show there is an $A \subseteq 2^{\omega}$ which is in $\Delta^0_{\alpha}$ but not in any $\Sigma^0_{\beta}$ for any $\beta<\alpha$. If $\alpha$ is a successor, say $\alpha=\beta+1$ then let $A\subseteq 2^{\omega}$ in $\Sigma^0_{\beta} \backslash \Pi^0_{\beta}$ and let $B \subseteq 2^{\omega}, B\in \Pi^0_{\beta} \backslash \Sigma^0_{\beta} $ . Define $C\subseteq 2^{\omega}$ such that we have $C(x) \leftrightarrow (x(0)=0 \wedge x(i+1)\in A \forall i) \vee (x(0)=1 \wedge x(i+1)\in B \forall i$. Then $C \in \Delta^0_{\alpha}$

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You meant $\alpha<\omega_1$ in that last part of the theorem statement? Since $\alpha<1\Rightarrow\alpha=0$... :-) –  Asaf Karagila Oct 9 '11 at 20:27
    
Yes, thank you :) –  Carlo Von Schnitzel Oct 9 '11 at 21:16
    
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 $\lbrace x\rbrace\times Y\subseteq A$. –  Asaf Karagila Oct 9 '11 at 22:36

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