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, ...} : #R##N#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 . #R##N#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^+ .
['Martin Goldstern', 'Saharon Shelah']
ANTICHAINS IN PRODUCTS OF LINEAR ORDERS