(D) Introduction to order types and ordinals Solutions to Exercises and Problems

1.3 We argue by contradiction. Thus we suppose that A is infinite and derive a contradiction. As a preliminary, observe that A is non-empty and hence has a ≤-least element, a first element. It also has a ≤∗-least element, a ≤-greatest element, a last element. For each element x let ↑x be the ≤-upper set generated by x, that is y ∈ ↑x ⇐⇒ x ≤ y for each y ∈ A. Consider the subset X ⊆ A given by x ∈ X ⇐⇒ ↑x is infinite for x ∈ A. Note that X is non-empty since, by hypothesis, the ≤-least element is in X. Note also that for x, y ∈ A we have y ≤ x =⇒ ↑x ⊆ ↑y Holds. Thus X is a ≤-lower section of A, that is y ≤ x ∈ X =⇒ y ∈ X for x, y ∈ A. The set X is non-empty, and A∗ is well-ordered, so X has a ≤∗-least element, a ≤-largest element. Let a be this element. Thus (i) ↑a is infinite (ii) ↑x is infinite ⇐⇒ x ≤ a