Antichains in products of linear orders

Martin Goldstern, Saharon Shelah

Published 1999-02-08Version 1

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^+ .