2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/780771. 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 pagesLogicGeneral TopologyPrimary 03E35; secondary 03E04, 06A05text