In Friedman [90] we constructed a Π1
이 논문은 Friedman(1990)이 제안한 문제를 해결하는 데 중점을 둔다. Friedman(1990)은 Π1
2-서클 R을 건설하여 0
T에 대한 명제는 ZFC의 서브테러리로서 사용된다. 또한 T는 consistent strength로 인접한 ineffectible cardinal을 포함한다. 본 논문에서는 Friedman(1990)의 건설 방법을 사용하여 타입 R의 Π1
2 formula가 유일하게 증명될 수 있는지 여부를 살펴보았다.
R은 finite sequences of 0's and 1's로 구성되어 있으며, p(i1, · · · , in+1)0가 포함된 generic class에서 결정된다. 본 논문에서는 T가 R의 Π1
2 formula가 유일하게 증명될 수 있는지 여부를 검증하였다.
한글 요약 끝
영어 요약:
Below is the English summary of the arXiv paper.
This paper addresses the open problem posed by Friedman (1990). Friedman (1990) constructed a $\Pi_1^2$-singleton R such that $0
The statement of T is used as a subtheory of ZFC. Also, T has consistency strength that is approximately that of ZFC+ there exists an ineffectible cardinal. We examine whether the $\Pi_1^2$ formula characterizing R is provably unique using the construction method of Friedman (1990).
R is determined by the $p(i_1,\dots,i_{n+1})^0$ where $p(i_1,\dots,i_{n+1})$ belongs to the generic class. We verify that T proves that the $\Pi_1^2$ formula characterizing R is provably unique.
English summary end.
In Friedman [90] we constructed a Π1
arXiv:math/9302201v1 [math.LO] 4 Feb 1993Provable Π12-SingletonsSy D. Friedman*M.I.T.In Friedman [90] we constructed a Π12-singleton R such that 0 An acceptable quess issuch a sequence (i1, · · · , in+1) where i1 is L-inaccessible and 1 ≤k < ℓ≤n −→ik ∈I(iℓ, · · · , in+1).Now we say that an acceptable guess (i1, · · · , in+1) is good if in addition I(i1, · · · , in+1)is stationary in i1. We refer to n as the length of the guess (i1, · · · , in+1).T is the theory ZFC+ There are arbitrarily long good guesses.T is a sub-theory of ZFC + 0# exists since any increasing sequence of Silver indiscernibles(i1, · · · , in+1), where n ≥1 and i1 is regular, is a good guess. (In fact I(i1, · · · , in+1)is CUB in i1 in this case.) Also note that T follows from the existence, for each n,of a cardinal K such that any function on n-tuples from K has a homogeneous setX containing an α such that X ∩α is stationary in α, together with n −2 largerordinals. (This is close in strength to an ineffable cardinal.) And if T is true thenit is true in L.Now recall that in Friedman [90] a Π12-singleton R is constructed so as to “kill” ac-ceptable guesses (i1, · · · , in+1) such that in+1 < (i+)L and p(i1, · · · , in+1)0 contra-dicts R. Here, p(i1, · · · , in+1) is a Σ1(L)-procedure which assigns a forcing conditionto the guess (i1, · · · , in+1) and p(i1, · · · , in+1)0 is the “real part” of p(i1, · · · , in+1),*Research supported by NSF contract # 9205530.1 2consisting of a function from (2<ω)<ω into perfect trees. R is in fact a set of finitesequences of finite sequences of 0’s and 1’s and is determined by the p(i1, · · · , in+1)0where p(i1, · · · , in+1) belongs to the generic class. A simple requirement that wemay impose on the procedure p(i1, · · · , in+1) is that p(i1, · · · , in+1) must decidewhich of the first n elements of (2<ω)<ω belongs to R, for some fixed (constructible)ω-listing of (2<ω)<ω.An acceptable guess (i1, · · · , in+1) is killed by adding a CUB subset to i1 disjointfrom I(i1, · · · , in+1).The Π12 formula characterizing R implies that R kills allacceptable guesses (i1, · · · , in+1) such that in+1 < (i+1 )L and p(i1, · · · , in+1) forcesa false membership fact about R. Now suppose T holds and that R ̸= S were bothsolutions to our Π12 formula. Choose n so that R and S differ on the membership ofone of the first n elements of (2<ω)<ω and let (i1, · · · , in+1) be a good guess. By aSkolem hull argument we may assume that in+1 < (i+1 )L. Then either R or S mustkill (i1, · · · , in+1) since p(i1, · · · , in+1) decides membership of the first n elementsof (2<ω)<ω. But goodness means that I(i1, · · · , in+1) is stationary, a contradiction.So T proves that our Π12 formula characterizing R has at most one solution.ReferenceFriedman [90]The Π12-Singleton Conjecture, Journal of the AMS. 출처: arXiv:9302.201 • 원문 보기