8
votes
1answer
89 views

Consistency strength of Turing measurability

This is probably well-known to recursion theorists, but as google didn't help me, I'll ask it here. Convention: All sets of reals in the following discussion are assumed to be closed under Turing ...
5
votes
0answers
121 views

How is the Kleene normal form theorem for $\Sigma^1_1$ relations proved in RCA0?

All of the following concerns Simpson's Subsystems of Second Order Arithmetic (2nd ed.). In the notes subsequent to lemmas VII.1.6 and VII.1.7 (pp. 245–246), Simpson remarks that both lemmas are ...
6
votes
1answer
301 views

What are $\Sigma _n^i$, $\Pi _n^i$ and $\Delta _n^i$?

Sometimes reading on wikipedia or in this site (and in very different context like topology, arithmetic and logic) I have found these symbols $\Sigma _n^i$, $\Pi _n^i$ and $\Delta _n^i$. They are ...
3
votes
1answer
142 views

Members of (lightface) Borel sets

I'm fairly certain this question has a very simple answer, and that I've learned it before; I just can't seem to remember it. Suppose I have a nonempty lightface Borel set $X\subseteq 2^\omega$. What ...
5
votes
2answers
254 views

Is there any generalization of the hyperarithmetical hierarchy using the analytical hierarchy to formulas belonging to third-order logic and above?

As I understand, hyperarithmetical sets are defined according to the analytical hierarchy, that is, second-order-logic formulas. There is a generalization of hyperarithmetic theory named α-recursion ...
6
votes
1answer
120 views

Terminal Paths in Kleene's O

I'm stuck on a problem in Sack's Higher Recursion Theory (#2.4)- any hints are welcome. He defines Kleene's O in the usual way, and the corresponding order $<_O$. A path through O is a linearly ...
6
votes
3answers
229 views

From lightface $\Sigma^1_1$ to boldface $\mathbf\Delta^1_1$

Fix some standard Polish space, e.g. Baire's space. It's a simple observation that every $\Delta^1_1$ is also $\mathbf\Delta^1_1$. It is the same observation that $\Sigma^1_1$ becomes ...
10
votes
2answers
180 views

How does Borelness overlap with definability, computability, or constructiveness?

Background: I am writing a short paper aimed at math undergrads and focused as narrowly as possible on Borel equivalence relations. So, e.g., I am not assuming familiarity with recursion theory and am ...
2
votes
2answers
127 views

Complexity of subset relation on Borel Sets

Fix $A, B \in \mathbf{\Delta}_{1}^{1}$ (i.e. they're Borel). Is the statement $A \subseteq B$ generally only $\mathbf{\Pi}^{1}_{1}$ (at best)? Of course, it's $\mathbf{\Pi}^{1}_{1}$ via $\forall x[x ...
4
votes
1answer
148 views

Projecting onto (lightface) Borel sets

Suppose $A \subseteq \omega^{\omega} \times \omega^{\omega}$ is Borel. If we project $A$ onto $\omega^\omega$, we get a $\mathbf{\Sigma^{1}_{1}}$ set $\{y: \exists x (y,x) \in A\}$. What if we project ...
3
votes
1answer
162 views

Proof of a Theorem in Gao's 'Invariant Descriptive Set Theory'

Theorem 1.7.5 on p.35 of Gao's Invariant Descriptive Set Theory reads Theorem 1.7.5 (Kleene) If $A\subseteq X \times \omega^{\omega}$ is $\Pi^{1}_{1}$ and $$x \in B \Longleftrightarrow \exists y ...
4
votes
1answer
107 views

Notation in Sacks' 'Higher Recursion Theory'

I'm having trouble with the notation in Sacks' Higher Recursion Theory. I've asked specific questions from page 4. Instead of reading my question in detail and trying to understand my confusion (which ...
2
votes
0answers
106 views

constructive ordinal and $\Delta^1_1$ predicate

Everything I know on this subject comes from Sacks book : "Higher recursion theory" Let $\mathcal{O^Y}$ be the set of codes for ordinals constructive in $Y$. We should have the result that $A ...
1
vote
2answers
63 views

Effective complexity of $\leq_T$

Remember that we say for $\alpha,\beta$ in $\omega^\omega$, that $\alpha\leq_T \beta$ if $\alpha$ is recursive in $\beta$. Is $\leq_T$ a $\Sigma^1_1$ set, as a subset of $\omega^\omega\times ...
7
votes
2answers
183 views

Complexity of the set of computable ordinals

According to http://en.wikipedia.org/wiki/Analytical_hierarchy The set of all natural numbers which are indices of computable ordinals is a $\Pi^1_1$ set which is not $\Sigma^1_1$. However, "the ...
5
votes
2answers
363 views

Does every infinite $\Sigma^1_1$ set have an infinite $\Delta^1_1$ subset?

The question is exactly that in the title: Does every infinite $\Sigma^1_1$ set of natural numbers have an infinite $\Delta^1_1$ subset? Some background: The lower-level analog of this question, Does ...