Pure Mathematics

## An Introduction to Set Theory by W. Weiss

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By W. Weiss

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Extra info for An Introduction to Set Theory

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Of course, the relation on the transitive set is the membership relation, which is extensional and well founded. This unexpected result leads to the important Mostowski Collapsing Theorem. Theorem 20. Let R be a well founded extensional relation on a set X. There is a unique transitive set M and a unique isomorphism h : X → M . Proof. Obtain f : X → δ directly from the previous theorem. By recursion on the ordinals we define for each β < δ a function hβ : f ← {β} → V such that hβ (y) = {hα (x) : α < β and x, y ∈ R}.

Show that ω + 1 is not a cardinal and that, in fact, each other cardinal is a limit ordinal. Theorem 23. The following are equivalent. 1. κ is a cardinal. 2. , |κ| = κ. 3. (∀α < κ)(¬∃ injection f : κ → α). 59 60 CHAPTER 7. CARDINALITY Proof. We prove the negations of each are equivalent: ¬(1) (∀x)[(∃ bijection g : x → κ) → (∃α < κ)(∃ bijection h : x → α)] ¬(2) ∃α < κ ∃ bijection f : κ → α ¬(3) ∃α < κ ∃ injection f : κ → α ¬(2) ⇒ ¬(1) Just take h = f ◦ g. ¬(1) ⇒ ¬(3) Just consider x = κ. ¬(3) ⇒ ¬(2) Suppose α < κ and f : κ → α is an injection.

If n1 ∩ S = ∅, Foundation gives us n2 ∈ n1 ∩ S with n2 ∩ (n1 ∩ S) = ∅. By transitivity, n2 ⊆ n1 so that n2 ∩ S = ∅. Thus, we always have some n ∈ S such that n ∩ S = ∅. For just such an n, choose m ∈ N with m ⊆ n, m = n, and m ∈ / n. Using Foundation, choose l ∈ n \ m such that l ∩ (n \ m) = ∅. Transitivity gives l ⊆ n, so we must have l ⊆ m. We have l = m since l ∈ n and m ∈ / n. Therefore we conclude that m \ l = ∅. Using Foundation, pick k ∈ m \ l such that k ∩ (m \ l) = ∅. Transitivity of m gives k ⊆ m and so we have k ⊆ l.