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Relating to the Turing degrees or Turing reducibility.
2
votes
1
answer
116
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$\Pi^0_2$ singleton forming minimal pair with $0''$
Is there a $\Pi^0_2$ singleton that forms a minimal pair with $0''$? That is, is there a set $X$ such that $X$ is the unique solution to $\forall x \exists y \phi(X|_y, x)$, $X$ and $0''$ are incompa …
2
votes
Double Posner-Robinson Join (or a cupping analog of minimal pair)
As pointed out in the comments, this is theorem 2 (relativized in theorem 3) in "Degrees joining to 0'" by Posner-Robinson. I summarize the method used below.
In the 1 degree version they use the fac …
1
vote
2
answers
129
views
Double Posner-Robinson Join (or a cupping analog of minimal pair)
Are there incompatible degrees $D_0, D_1 <_T 0'$ such that for all $X$ if $D_0 \oplus X \equiv_T D_1 \oplus X \equiv_T 0'$ then $X \equiv_T 0'$? So kinda like a cupping analog of a minimal pair.
To p …
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 $X' \geq_T 0''$.
1
vote
1
answer
107
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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 …
1
vote
0
answers
37
views
Are the $\omega$-generic arithmetic degrees downward closed
A degree is $\alpha$-generic if it has representative that is $\alpha$-generic. Are the $\omega$-generic arithmetic degrees (i.e. the degree structure induced by arithmetic reproducibility) downward …
2
votes
Accepted
Does every arithmetic degree below $0^\omega$ have a representative computable in $0^\omega$?
Ok, I'm pretty sure the answer is no to the basic question. We build a set $A$ to satisfy the following conditions.
$P_e: A \neq \phi_i(0^{n})$
$R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X …
3
votes
1
answer
91
views
Does every arithmetic degree below $0^\omega$ have a representative computable in $0^\omega$?
Suppose that $A \leq_a 0^\omega$ (i.e. $A$ is arithmetic in $0^\omega$) does there exist $\widehat{A} \equiv_a A$ with $\widehat{A} \leq_T 0^\omega$ [1]?
More generally, say that a set $X$ is aT-compl …
2
votes
Incompatible degrees $a,b$ s.t. $x < a$ implies $x \leq b$
Just in case anyone else is looking what was nagging me about having seen a construction for this it's essentially the result of the minimal upper bound construction.
Let $a$ be any degree that's not …
2
votes
3
answers
137
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Incompatible degrees $a,b$ s.t. $x < a$ implies $x \leq b$
Are there incompatible Turing degrees $a,b$ s.t any degree computable in $a$ either computes $a$ or is computed by $b$?
Obviously, if $a$ was above $b$ then $a$ would be a strong minimal cover of $b$. …