18. Large cardinals

(i) F is an ultrafilter.
(ii) F is maximal in the partially ordered set of all proper filters (under ⊆).
Proof. (i)⇒(ii): Assume (i), and suppose that G is a filter with F ⊂ G . Choose
Y ∈ G \F. Since Y /∈ F, we must have X\Y ∈ F ⊆ G . So Y, X\Y ∈ G , hence
∅ = Y ∩ (X\Y ) ∈ G , and so G is not proper.
(ii)⇒(i): Assume (ii), and suppose that Y ⊆ X, with Y /∈ F; we want to show that
X\Y ∈ F. Let
G = {Z ⊆ X : Y ∩ W ⊆ Z for some W ∈ F }.
Clearly G is a filter on X, and F ⊆ G . Moreover, Y ∈ G \F. It follows that G is not
proper, and so ∅ ∈ G . Thus there is a W ∈ F such that Y ∩ W = ∅. Hence W ⊆ X\Y ,
and hence X\Y ∈ F, as desired.
Theorem 18.18. For any infinite set X there is a nonprincipal ultrafilter on X. Moreover,
if A is any collection of subsets of X with fip, then A can be extended to an ultrafilter.
Proof. First we show that the first assertion follows from the second. Let A be the
collection of all cofinite subsets of X—the subsets whose complements are finite. A has
fip, since if B is a finite subset of A , then X\ ⋂ B = ⋃ Y ∈B
(X\B) is finite. By the second
assertion, A can be extended to an ultrafilter F. Clearly F is nonprincipal.
To prove the second assertion, let A be a collection of subsets of X with fip, and let
C be the collection of all proper filters on X which contain A . Clearly the filter generated
by A is proper, so C ≠ ∅. We consider C as a partially ordered set under inclusion.
Any subset D of C which is a chain has an upper bound in C , namely ⋃ D, as is easily
checked. So by Zorn’s lemma C has a maximal member F. By Proposition 18.16, F is an
ultrafilter.
• Let X be an infinite set, and let κ be an infinite cardinal. An ultrafilter F on X is κ-
complete iff for any A ∈ [F] 0 let Z α =
( ⋂ β