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Descriptive Set Theory is the study of definable subsets of Polish spaces, where definable is taken to mean from the Borel or projective hierarchies. Other topics include infinite games and determinacy, definable equivalence relations and Borel reductions between them, Polish groups, and effective descriptive set theory.
0
votes
Accepted
A partial relativization of Gandy's basis theorem
Hmmm, it seems the answer to the question is no.
Fix any nonhyperarithmetic real $x$ so that the $\Sigma^1_1(x)$ set $$A_x=\{y\mid \forall n \in \mathscr{O} ( x\not\leq_T y^{(|n|)})\}$$ is not empty, …
4
votes
Vitali Sets vs Bernstein Sets...
For your second definition of Vitali set, I have a weak partial answer. Namely the existence of a Bernstein set does not imply the existence of a $T$-Vitali set. The answer can be found in logic blog …
4
votes
Accepted
Analytic uniformization
There is an arithmetical set $A\subseteq 2^{<\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(x|n,\sigma)\in A\}$ is an $x$-recursive tree which has an i …
3
votes
$\Delta^{1}_{2}$ and degrees of constructibility $\textbf{on sets}$
Here is a partial positive answer: i.e. either $A$ or $\bar{A}$ contain elements of all degrees of constructibility of reals.
Given an infinite set of numbers $x$, let $f:\omega\to \omega\cup \{x\} $ …
2
votes
A Borel perfectly everywhere surjective function on the Cantor set
I think that recursion theory gives a clearer way to answer the question. If $f$ is a Borel function, then it is a hyperarithmetic reduction relative to a real, say $x$. Then fix any "regular" forcin …
11
votes
Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets
Here is another example from recursion theory:
Take a chain $\{x_{\alpha}\}_{\alpha<\omega_1}$ from Turing degrees.
For each $\alpha<\omega_1$, let $A_{\alpha}$ be the collection of the reals neithe …
6
votes
Borel cross section
There is a recursion theory method which is quite similar to Samuel's.
By a classical recursion theory result, there is a recursive functional $F$ so that for any $x\in \omega^{\omega}$, $F(x)$ code …
4
votes
Definition of HYP in $L_{\omega_1^{CK}}[a]$?
The following is not the answer of your question. But I think it is what you really want.
I guess you might be figuring out Leo's proof of McLaughlin's conjecture and his answer to Question 65 in Har …
2
votes
Regularity properties of Turing-invariant and arbitrary sets of reals
Here is not an answer but an example:
Let PSP be the statement that every uncountable set of reals has a perfect subset and TPSP that every uncountable set of Turing degrees has a perfect subset.
Th …
1
vote
Accepted
The measure of ideals generated by random reals
The question has a negative answer. The technique is essentially due to Jockusch and Posner.
Proof: Let $x$ be a real in which every constructible real is recursive. Now $$A=\{r\mid r\mbox{ is Mar …
1
vote
Borel hierarchy and tail sets
Here is an example: Let $A=\{0,1\}$ and for any $x\in \{0,1\}^{\omega}$, let $n<_x m$ if $x(2^n\cdot 3^m)=1$. For any countable ordinal $\alpha$, let $x\in U_{\alpha}$ if there is some $l$ so that $ …
4
votes
Accepted
$\Pi^0_2$ singleton forming minimal pair with $0''$
Maybe I should give a more detailed answer.
Harrington proved (or claimed) the following result in his handwritten draft.
Theorem There is a $\Pi^0_2$-singleton $x$ so that $\forall n<\omega (x^{(n)} …
1
vote
Hyperarithmetically least elements in $\Pi^1_1$ sets
It is well known that there is a nonempty $\Pi^0_1$ set $A\subseteq \omega^{\omega}$ containing no hyperarithmetic member.(Considering a $\Sigma^1_1$ set without a hyperarithmetic member, it is a proj …
4
votes
A compactness property for Borel sets
Here is a even simpler example.
Let $\{x_{\alpha} \}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}=\{y\mid y\geq_T x_{\alpha}\}$. Each $B_{\a …
4
votes
Accepted
Degree of unsolvability of finding a open approximation to a Borel set, given its Borel code
I think this is a well known fact.
For example, let $\alpha$ be a recursive ordinal and $B=\{g\mid g\mbox{ is a }0^{(\alpha)}\mbox{-generic real}\}$. Then $B$ is a hyperarithmetic set and so has a re …