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Hk−1 , hk+1 , . . hn+2 are arbitrary elements from H, then there are fewer than ℵα elements h ∈ H such that (h1 , . . , hk−1 , h, hk+1 , . . hn+2 ) ∈ Hk . 16. (Cantor’s inequality) For any κ we have 2κ > κ. 17. (K¨ onig’s inequality) If ρi < κi for all i ∈ I, then ρi < i∈I κi . i∈I 18. If the set of cardinals {κξ }ξ<θ , 0 < κξ < κ is cofinal with κ, then ξ<θ κξ > κ. 19. If κ is infinite, κ = ξ
4. If in a partially ordered set all antichains have at most k < ∞ elements, then the set is the union of k chains. 5. There is a partially ordered set in which all chains are finite, still the set is not the union of countably many antichains. 6. There is a partially ordered set in which all antichains are finite, still the set is not the union of countably many chains. 56 Chapter 11 : Partially ordered sets Problems 7. If in a partially ordered set all chains are finite and all antichains are countable, then the set is countable.
4. α is divisible from the right by 2 and 3 if and only if it is divisible from the right by 6. Is the same true for divisibility from the left? 5. α is divisible from the right by ω + 2 and by ω + 3 if and only if it is divisible from the right by ω + 6. 6. Every ordinal α has only a finite number of right divisors. Is the same true of left divisors? What if α is a successor ordinal? 7. If α and β are right divisors of γ ≥ 1, then either a) α divides β from the right, or b) β divides α from the right, or c) α = ξ +p, β = ξ +q, where ξ is a limit ordinal or 0, and p, q are positive natural numbers.