Where pigeonhole principles meet Koenig lemmas

Belanger, David; Chong, Chai Tai; Wang, Wei; Wong, Tin Lok; Yue, Yang

doi:10.1090/tran/8494

Cited by 2 publications

(38 citation statements)

References 24 publications

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“…Over BΣ 2 , one basic issue is whether the first-order consequences of RCA * 0 + RT We have no strong reasons to believe that the answer is "yes." However, it should be pointed out that, since RCA 0 + RT 2 2 proves BΣ 2 , answering "no" would involve constructing a model of IΣ 1 + ¬BΣ 2 that expands to a model of BΣ 0 1 + ¬IΣ 0 1 -in the terminology of [23], a model of IΣ 1 + ¬BΣ 2 that is not always semiregular. The existence of such a model itself seems to be open.…”

Section: Proof It Has Already Been Mentioned That Bς + Exp Implies Gp...mentioning

confidence: 99%

“…Then, in Section 5, we take a look at the question whether our results say anything about Ramsey's Theorem for pairs over RCA 0 . We consider a principle that can be viewed as a "jumped version" of RT 2 2 , and we show that it is not Π 5 -conservative over RCA 0 + BΣ 0 2 . We also show that the most obvious sentence witnessing the lack of conservativity is unprovable in RCA 0 + RT 2 2 .…”

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confidence: 99%

“…We consider a principle that can be viewed as a "jumped version" of RT 2 2 , and we show that it is not Π 5 -conservative over RCA 0 + BΣ 0 2 . We also show that the most obvious sentence witnessing the lack of conservativity is unprovable in RCA 0 + RT 2 2 . However, the proof of unprovability, which is based on a possibly unexpected technique (proof speedup), no longer works for slightly weaker sentences.…”

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confidence: 99%

“…Recent work of Belanger [3] has demonstrated that the study of reverse mathematics over RCA * 0 is relevant to the traditional RCA 0 framework as well. In fact, a large part of our original motivation for studying Ramsey's Theorem over RCA * 0 was the desire to understand whether it can help in understanding RT 2 2 over RCA 0 . The jury is still out on that.…”

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confidence: 99%

“…Proof. Fix n ≥ 3 and let R n be as in (2). We first argue that for every M |= R n there is a family of sets…”

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confidence: 99%

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How Strong Is Ramsey’s Theorem if Infinity Can Be Weak?

Kołodziejczyk¹,

Kowalik²,

Yokoyama³

2022

J. symb. log.

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We study the first-order consequences of Ramsey’s Theorem for k-colourings of n-tuples, for fixed $n, k \ge 2$ , over the relatively weak second-order arithmetic theory $\mathrm {RCA}^*_0$ . Using the Chong–Mourad coding lemma, we show that in a model of $\mathrm {RCA}^*_0$ that does not satisfy $\Sigma ^0_1$ induction, $\mathrm {RT}^n_k$ is equivalent to its relativization to any proper $\Sigma ^0_1$ -definable cut, so its truth value remains unchanged in all extensions of the model with the same first-order universe. We give a complete axiomatization of the first-order consequences of $\mathrm {RCA}^*_0 + \mathrm {RT}^n_k$ for $n \ge 3$ . We show that they form a non-finitely axiomatizable subtheory of $\mathrm {PA}$ whose $\Pi _3$ fragment coincides with $\mathrm {B} \Sigma _1 + \exp $ and whose $\Pi _{\ell +3}$ fragment for $\ell \ge 1$ lies between $\mathrm {I} \Sigma _\ell \Rightarrow \mathrm {B} \Sigma _{\ell +1}$ and $\mathrm {B} \Sigma _{\ell +1}$ . We also give a complete axiomatization of the first-order consequences of $\mathrm {RCA}^*_0 + \mathrm {RT}^2_k + \neg \mathrm {I} \Sigma _1$ . In general, we show that the first-order consequences of $\mathrm {RCA}^*_0 + \mathrm {RT}^2_k$ form a subtheory of $\mathrm {I} \Sigma _2$ whose $\Pi _3$ fragment coincides with $\mathrm {B} \Sigma _1 + \exp $ and whose $\Pi _4$ fragment is strictly weaker than $\mathrm {B} \Sigma _2$ but not contained in $\mathrm {I} \Sigma _1$ . Additionally, we consider a principle $\Delta ^0_2$ - $\mathrm {RT}^2_2$ which is defined like $\mathrm {RT}^2_2$ but with both the $2$ -colourings and the solutions allowed to be $\Delta ^0_2$ -sets rather than just sets. We show that the behaviour of $\Delta ^0_2$ - $\mathrm {RT}^2_2$ over $\mathrm {RCA}_0 + \mathrm {B}\Sigma ^0_2$ is in many ways analogous to that of $\mathrm {RT}^2_2$ over $\mathrm {RCA}^*_0$ , and that $\mathrm {RCA}_0 + \mathrm {B} \Sigma ^0_2 + \Delta ^0_2$ - $\mathrm {RT}^2_2$ is $\Pi _4$ - but not $\Pi _5$ -conservative over $\mathrm {B} \Sigma _2$ . However, the statement we use to witness failure of $\Pi _5$ -conservativity is not provable in $\mathrm {RCA}_0 +\mathrm {RT}^2_2$ .

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Section: Proof It Has Already Been Mentioned That Bς + Exp Implies Gp...mentioning

confidence: 99%

mentioning

confidence: 99%

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confidence: 99%

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confidence: 99%

“…Proof. Fix n ≥ 3 and let R n be as in (2). We first argue that for every M |= R n there is a family of sets…”

mentioning

confidence: 99%

See 3 more Smart Citations

How Strong Is Ramsey’s Theorem if Infinity Can Be Weak?

Kołodziejczyk¹,

Kowalik²,

Yokoyama³

2022

J. symb. log.

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THE REVERSE MATHEMATICS OF ${\mathsf {CAC\ FOR\ TREES}}$

CERVELLE¹,

GAUDELIER²,

Patey

2023

J. symb. log.

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${\mathsf {CAC\ for\ trees}}$ is the statement asserting that any infinite subtree of $\mathbb {N}^{<\mathbb {N}}$ has an infinite path or an infinite antichain. In this paper, we study the computational strength of this theorem from a reverse mathematical viewpoint. We prove that ${\mathsf {CAC\ for\ trees}}$ is robust, that is, there exist several characterizations, some of which already appear in the literature, namely, the statement $\mathsf {SHER}$ introduced by Dorais et al. [8], and the statement $\mathsf {TAC}+\mathsf {B}\Sigma ^0_2$ where $\mathsf {TAC}$ is the tree antichain theorem introduced by Conidis [6]. We show that ${\mathsf {CAC\ for\ trees}}$ is computationally very weak, in that it admits probabilistic solutions.

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Where pigeonhole principles meet Koenig lemmas

Cited by 2 publications

References 24 publications

How Strong Is Ramsey’s Theorem if Infinity Can Be Weak?

How Strong Is Ramsey’s Theorem if Infinity Can Be Weak?

THE REVERSE MATHEMATICS OF ${\mathsf {CAC\ FOR\ TREES}}$

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