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Used for questions about REA (aka CEA), sets in computability theory where REA stands for recursively enumerable and above.
2
votes
0
answers
40
views
Uniformity of splitting for n-REA degrees
In "A SPLITTING THEOREM FOR n−REA DEGREES" Shore and Slaman extend the following result of Sacks
If $C$ is r.e., $D \leq_T C$ and $D, C \not\leq_T 0$ then there are sets $C_0, C_1$ such that $C \equ …
2
votes
1
answer
99
views
Generality of construction for $\omega$-REA arithmetic degrees
So a common method used to construct non-zero $\omega$-REA arithmetic degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)
$$\tag{1} J …
0
votes
Accepted
Generality of construction for $\omega$-REA arithmetic degrees
So after some careful thought I'm pretty sure it is fully general. Let $A$ be a non-arithmetic $\omega$-REA set and $K(X)$ some $\omega$-REA operator satisfying 1 and 2 (see Odifreddi volume 2 XIII.3 …
1
vote
1
answer
66
views
Computable in $\omega$-REA degree but not double jump of finitely many columns
Suppose that $A$ is an $\omega$-REA set (so $A^{[n]}$ is r.e. in the prior columns). It is a well-known result that if each column of $A$ is computable then $A \leq_T \emptyset^2$ ($\emptyset^2$ can …
2
votes
1
answer
71
views
Arithmetic non-trivial 2-l.u.b
Remember a degree $\mathbb{d}$ is the $n$-lub of $\mathbb{c}_j$ in the Turing degrees if it is the least element (not merely a minimal element) set of $\mathbb{c}^{(n)}$ such that $\mathbb{c}$ comput …
0
votes
Accepted
Computable in $\omega$-REA degree but not double jump of finitely many columns
Ok, so unfortunately the answer is no. I've attached a proof that builds a $\omega$-REA set such that $A^{[\leq n]}$ is low for all $n$ yet $A$ computes $\emptyset'' \oplus X$ where $X$ is a natural …
0
votes
Accepted
Arithmetic non-trivial 2-l.u.b
Yes, it turns out that there is such an arithmetic (indeed 3-REA) non-trivial 2-lub. In fact, $\emptyset'''$ is such a degree. Consider the construction I give here in answer to this question.
This …
7
votes
1
answer
166
views
2-REA PA degrees
Remember that an n-REA set is a set of the form $A_0 \oplus A_1 \oplus \cdots \oplus A_n$ with $A_n$ relatively r.e. in $A_m, m<n$ (so $A_0$ is r.e.) and that a degree is PA just if it computes a path …
1
vote
1
answer
107
views
Double Hop Inversion Theorem
The hop $H_e$ is defined by $H_e(X) = X \oplus W_e^{X}$. A 2-REA operator (or double hop) $J_{\langle e,i\rangle}$ is defined by $J_{\langle e,i\rangle}(X) = H_e(H_i(X))$
By a famous result from Pse …
2
votes
Accepted
Double Hop Inversion Theorem
Ohh, I think I was being dumb. There is a 2-REA operator $J$ such that $J(X) <_T X'$ isn't of degree r.e. in $X$. Since $0''$ is of r.e. degree in every $X < 0''$ with $X' \geq_T 0''$.
2
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0
answers
92
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Direct construction of an arithmetically high degree below $0^{(\omega)}$
The existence of a high arithmetic degree (meaning the degrees induced by the notion of relative arithmetic definability) below $0^\omega$ can be established by using Harrington/Simpson's constructio …