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Is every degree above ${bf 0''}$ PA over something close to itself?

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2

$begingroup$

The low basis theorem says that there are PA degrees which are low - that is, which satisfy ${bf a'}={bf 0'}$. Appropriately relativized, given a degree ${bf a}$ there is a degree ${bf b}$ "not much bigger than" ${bf a}$ which is PA over ${bf a}$. A natural question at this point is whether the "dual" holds: if ${bf b}$ is PA in the first place, must ${bf b}$ be PA over some ${bf a}$ which is "close to" ${bf a}$? One natural way to phrase this question would be: if ${bf b}$ is PA, then must ${bf b}$ be PA over some ${bf a}$ with ${bf a'}={bf b'}$?

It turns out that (quite surprisingly to me) the answer is no: the degree ${bf 0'}$ is PA but not PA over anything not low.

Say that a degree ${bf d}$ is relatively efficiently PA if there is some ${bf b}le_T{bf d}$ such that ${bf d}$ is PA over ${bf b}$ and ${bf d}$ is low over ${bf b}$ (that is, ${bf b'}={bf d'}$). By the observation cited above, ${bf 0'}$ is not relatively efficiently PA. However, in a precise sense "most" Turing degrees are relatively efficiently PA: the relativized low basis theorem tells us that the set of relatively efficiently PA degrees is unbounded, and so Martin's cone theorem (+ the fact that "relatively efficiently PA" is a sufficiently simple property) tells us that the set of relatively efficiently PA degrees contains a cone.

My question is roughly when this happens - that is, what might be a reasonable base for such a cone. Specifically:

Is every degree $ge_T{bf 0''}$ relatively efficiently PA?

logic computability

share|cite|improve this question

edited Mar 20 at 1:14

Noah Schweber

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

$endgroup$

add a comment | 

2

$begingroup$

The low basis theorem says that there are PA degrees which are low - that is, which satisfy ${bf a'}={bf 0'}$. Appropriately relativized, given a degree ${bf a}$ there is a degree ${bf b}$ "not much bigger than" ${bf a}$ which is PA over ${bf a}$. A natural question at this point is whether the "dual" holds: if ${bf b}$ is PA in the first place, must ${bf b}$ be PA over some ${bf a}$ which is "close to" ${bf a}$? One natural way to phrase this question would be: if ${bf b}$ is PA, then must ${bf b}$ be PA over some ${bf a}$ with ${bf a'}={bf b'}$?

It turns out that (quite surprisingly to me) the answer is no: the degree ${bf 0'}$ is PA but not PA over anything not low.

Say that a degree ${bf d}$ is relatively efficiently PA if there is some ${bf b}le_T{bf d}$ such that ${bf d}$ is PA over ${bf b}$ and ${bf d}$ is low over ${bf b}$ (that is, ${bf b'}={bf d'}$). By the observation cited above, ${bf 0'}$ is not relatively efficiently PA. However, in a precise sense "most" Turing degrees are relatively efficiently PA: the relativized low basis theorem tells us that the set of relatively efficiently PA degrees is unbounded, and so Martin's cone theorem (+ the fact that "relatively efficiently PA" is a sufficiently simple property) tells us that the set of relatively efficiently PA degrees contains a cone.

My question is roughly when this happens - that is, what might be a reasonable base for such a cone. Specifically:

Is every degree $ge_T{bf 0''}$ relatively efficiently PA?

logic computability

share|cite|improve this question

edited Mar 20 at 1:14

Noah Schweber

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

$endgroup$

add a comment | 

$begingroup$

The low basis theorem says that there are PA degrees which are low - that is, which satisfy ${bf a'}={bf 0'}$. Appropriately relativized, given a degree ${bf a}$ there is a degree ${bf b}$ "not much bigger than" ${bf a}$ which is PA over ${bf a}$. A natural question at this point is whether the "dual" holds: if ${bf b}$ is PA in the first place, must ${bf b}$ be PA over some ${bf a}$ which is "close to" ${bf a}$? One natural way to phrase this question would be: if ${bf b}$ is PA, then must ${bf b}$ be PA over some ${bf a}$ with ${bf a'}={bf b'}$?

It turns out that (quite surprisingly to me) the answer is no: the degree ${bf 0'}$ is PA but not PA over anything not low.

Say that a degree ${bf d}$ is relatively efficiently PA if there is some ${bf b}le_T{bf d}$ such that ${bf d}$ is PA over ${bf b}$ and ${bf d}$ is low over ${bf b}$ (that is, ${bf b'}={bf d'}$). By the observation cited above, ${bf 0'}$ is not relatively efficiently PA. However, in a precise sense "most" Turing degrees are relatively efficiently PA: the relativized low basis theorem tells us that the set of relatively efficiently PA degrees is unbounded, and so Martin's cone theorem (+ the fact that "relatively efficiently PA" is a sufficiently simple property) tells us that the set of relatively efficiently PA degrees contains a cone.

My question is roughly when this happens - that is, what might be a reasonable base for such a cone. Specifically:

Is every degree $ge_T{bf 0''}$ relatively efficiently PA?

logic computability

share|cite|improve this question

edited Mar 20 at 1:14

Noah Schweber

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

$endgroup$

share|cite|improve this question

edited Mar 20 at 1:14

Noah Schweber

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

share|cite|improve this question

edited Mar 20 at 1:14

Noah Schweber

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294

asked Mar 20 at 1:08

Noah SchweberNoah Schweber

128k10152294