Math 4200
Theorem: If F is an ordered field then F must contain infinitely many elements.
Proof:
Let f1 = 1 , and for each n ≥ 2 , let f n +1 = f n + 1 . We will show that ∀i, j ∈ J , f i < f j
whenever i < j . Suppose ∃i and ∃j such that i < j and f j ≤ f i . Let
S = { n ∈ J : i < n and f n ≤ f i } . S ≠ φ because j ∈ S . By mathematical induction, S
contains a least element, call it k. Consider f k −1 . Since k − 1 < k , we have k − 1 ∉ S
and so f i < f k −1 . Since 0 < 1 , we have f i + 0 < f k −1 + 1 . That is, f i < f k , a
contradiction. So S = φ and ∀i, j ∈ J , f i < f j whenever i < j . Therefore, F contains
infinitely many elements.