Notes on the theory of cardinals
نویسنده
چکیده
Lemma 3 On a well-ordered set, a strictly monotone function is extensive. Proof. Assume that S = {x ∈ X | x > f(x)} 6= ∅. Then, let a be the least element of S (X is well ordered). Hence, a > f(a). By strict monotony, f(a) > f(f(a)). Therefore, f(a) ∈ S and a ≤ f(a). Contradiction. Lemma 4 If f is a monotone injection from α to β, then α ≤ β. Proof. Since α is well ordered, f is extensive. Hence, for all x < α, x ≤ f(x) < β. If β < α, then β < β. Contradiction.
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