18. Large cardinals

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contradiction. So |P| = κ. Theorem <strong>18.</strong>20. Suppose that κ is the least infinite cardinal such that there is a nonprincipal σ-complete ultrafilter F on κ. Then F is κ-complete. Proof. Assume the hypothesis, but suppose that F is not κ-complete. So there is a A ∈ [F]
{α < κ : f α (β) = 0} and {α < κ : f α (β) = 1} is in U, so we can let ε(β) ∈ 2 be such that {α < κ : f α (β) = ε(β)} ∈ U. Then ⋂ {α < κ : f α (β) = ε(β)} ∈ U; β
Page 1 and 2: 18. Large cardinals The study, or u
Page 3 and 4: Assume the linear order property, a
Page 5 and 6: Now let α < κ; we show that B has
Page 7 and 8: so since |B| = κ it follows that |
Page 9 and 10: By induction, |F α | ≤ κ for al
Page 11 and 12: Lemma 18.9. Let A be a set of infin
Page 13 and 14: Proof. Suppose not, and let X = {b
Page 15 and 16: Clearly (V κ , ∈, U) |= σ. Supp
Page 17: (i) F is an ultrafilter. (ii) F is
Page 21 and 22: • I0 • huge • Vopěnka • ex
Page 23: {X ⊆ A : µ(X) = µ(A)} is a κ-c

18. Large cardinals

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