Harrington’s Principle in Higher Order Arithmetic

Abstract

Let Z2, Z3, and Z4 denote 2nd, 3rd, and 4th order arithmetic, respectively. We let Harrington’s Principle, HP, denote the statement that there is a real x such that every x-admissible ordinal is a cardinal in L. The known proofs of Harrington’s theorem “$Det\left( {{\rm{\Sigma }}_1^1} \right)$ implies 0♯ exists” are done in two steps: first show that $Det\left( {{\rm{\Sigma }}_1^1} \right)$ implies HP, and then show that HP implies 0♯ exists. The first step is provable in Z2. In this paper we show that Z2 + HP is equiconsistent with ZFC and that Z3 + HP is equiconsistent with ZFC + there exists a remarkable cardinal. As a corollary, Z3 + HP does not imply 0♯ exists, whereas Z4 + HP does. We also study strengthenings of Harrington’s Principle over 2nd and 3rd order arithmetic.

The strong reflecting property and Harrington's Principle

“…So the consistency of (1) implies the consistency of . By [, Theorem 3.2], implies that is remarkable”. By Proposition , is inaccessible in .…”

“…In , we define as the minimal iterable mouse and prove in that is equivalent to “ exists”. Theorem proves that these two definitions of are equivalent in .…”

Current Research on Gödel’s Incompleteness Theorems

“…• Section 0.7 on Incompleteness in Nested Multiply Recursive Arithmetic and Two Quantifier Arithmetic. Related to Friedman's work, Cheng [21,24] gives an example of concrete mathematical theorems based on Harrington's principle which is isolated from the proof of the Harrington's Theorem (the determinacy of Σ 1 1 games implies the existence of zero sharp), and shows that this concrete theorem saying that Harrington's principle implies the existence of zero sharp is expressible in Second-Order Arithmetic, not provable in Second-Order Arithmetic or Third-Order Arithmetic, but provable in Fourth-Order Arithmetic (i.e., the minimal system in higher-order arithmetic to prove this concrete theorem is Fourth-Order Arithmetic).…”

“…Eighth, we discuss some variants of R in the same language as L(R) = {0, ...,n, ..., + , × , ≤}. The theory R 0 is no longer essentially undecidable in the same language as R. 24 In fact, whether R 0 is essentially undecidable depends on the language of R 0 : if L(R 0 ) = {0,S, + , × , ≤} with ≤ defined in terms of +, then R 0 is essentially undecidable (Cobham first observed that R is interpretable in R 0 in the same language {0,S, + ,×}, and hence R 0 is essentially undecidable (see [65,127]). The theory R 1 is essentially undecidable since R is interpretable in R 1 (see [65, p. 62]).…”

Forcing a set model of + Harrington's Principle

“…In this step, we shoot a club C through S via Baumgartner's forcing P B S such that if η is the limit point of C and 4 We still work in…”