Antichains in products of linear orders

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1. For many regular cardinals lambda (in particular, for all successors of singular strong limit cardinals, and for all successors of singular omega-limits), for all n in {2,3,4, ...} : There is a linear order L such that L^n has no (incomparability-)antichain of cardinality lambda, while L^{n+1} has an antichain of cardinality lambda . 2. For any nondecreasing sequence (lambda2,lambda3, ...) of infinite cardinals it is consistent that there is a linear order L such that L^n has an antichain of cardinality lambda_n, but not one of cardinality lambda_n^+ .
9 pages

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Logic, General Topology, Primary 03E35; secondary 03E04, 06A05

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http://salesiana.dossiersoluciones.com/handle/123456789/78077

Consulte el texto completo en el siguiente enlace:

https://arxiv.org/abs/math/9902054
 http://arxiv.org/abs/math/9902054
 Order 19 No. 3 (2002) 213--222

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arXiv