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  1. A tail Cone version of the halpern–läuchli theorem at a large cardinal.Jing Zhang - 2019 - Journal of Symbolic Logic 84 (2):473-496.
    The classical Halpern–Läuchli theorem states that for any finite coloring of a finite product of finitely branching perfect trees of height ω, there exist strong subtrees sharing the same level set such that tuples in the product of the strong subtrees consisting of elements lying on the same level get the same color. Relative to large cardinals, we establish the consistency of a tail cone version of the Halpern–Läuchli theorem at a large cardinal (see Theorem 3.1), which, roughly speaking, deals (...)
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  • The halpern–läuchli theorem at a measurable cardinal.Natasha Dobrinen & Dan Hathaway - 2017 - Journal of Symbolic Logic 82 (4):1560-1575.
    Several variants of the Halpern–Läuchli Theorem for trees of uncountable height are investigated. Forκweakly compact, we prove that the various statements are all equivalent, and hence, the strong tree version holds for one tree on any weakly compact cardinal. For any finited≥ 2, we prove the consistency of the Halpern–Läuchli Theorem ondmany normalκ-trees at a measurable cardinalκ, given the consistency of aκ+d-strong cardinal. This follows from a more general consistency result at measurableκ, which includes the possibility of infinitely many trees, (...)
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  • Partitions of large Rado graphs.M. Džamonja, J. A. Larson & W. J. Mitchell - 2009 - Archive for Mathematical Logic 48 (6):579-606.
    Let κ be a cardinal which is measurable after generically adding ${\beth_{\kappa+\omega}}$ many Cohen subsets to κ and let ${\mathcal G= ( \kappa,E )}$ be the κ-Rado graph. We prove, for 2 ≤ m < ω, that there is a finite value ${r_m^+}$ such that the set [κ] m can be partitioned into classes ${\langle{C_i:i (...)
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  • Models without indiscernibles.Fred G. Abramson & Leo A. Harrington - 1978 - Journal of Symbolic Logic 43 (3):572-600.
    For T any completion of Peano Arithmetic and for n any positive integer, there is a model of T of size $\beth_n$ with no (n + 1)-length sequence of indiscernibles. Hence the Hanf number for omitting types over T, H(T), is at least $\beth_\omega$ . (Now, using an upper bound previously obtained by Julia Knight H (true arithmetic) is exactly $\beth_\omega$ ). If T ≠ true arithmetic, then $H(T) = \beth_{\omega1}$ . If $\delta \not\rightarrow (\rho)^{ , then any completion of (...)
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