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Asaf Karagila
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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.

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 $X$)

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.

Source Link
Asaf Karagila
  • 39.7k
  • 8
  • 135
  • 283

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

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 $X$)