紧致性,内插和弗里德曼的第三个问题

Daniele Mundici
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引用次数: 8

摘要

让Robinson的相合定理在逻辑L中成立,则L满足所有通常的插值性和可定义性以及可计数紧性,只要L相当小。后一种假设可以通过使用特殊的集合理论假设来削弱或消除。因此,如果Robinson的一致性定理在L中成立,则(i) L是可数紧的,如果它的Löwenheim个数为<μ0 =最小不可数可测基数;(ii)如果ω是唯一可测量的基数,则L是可数紧的,或者L的理论表征了每一个结构直到同构。作为推论,通过证明在L∞ω和L∞∞之间没有逻辑L严格满足内插(或鲁滨逊一致性),除非k -初等等价与同构重合,给出了H. Friedman第三个问题的部分答案。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Compactness, interpolation and Friedman's third problem

Let Robinson's consistency theorem hold in logic L: then L will satisfy all the usual interpolation and definability properties, together with coutable compactness, provided L is reasonably small. The latter assumption can be weakened ro removed by using special set-theoretical assumptions. Thus, if Robinson's consistency theorem holds in L, then (i) L is countably compact if its Löwenheim number is < μ0 = the smallest uncountable measurable cardinal; (ii) if ω is the only measurable cardinal, L is countably compact, or the theories of L characterize every structure up to isomorphism. As a corollary, a partial answer is given to H. Friedman's third problem, by proving that no logic L strictly between Lω and L∞∞ satisfies interpolation (or Robinson's consistency), unless K-elementary equivalence coincides with isomorphism.

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