# Least Upper Bound Property ⇒ Archimedean Principle

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Hello! I was wondering if this proof was correct? Thanks in advance!

Given: A totally ordered field, ##\mathbb{F}##.
Claim: Least Upper Bound Property (l.u.b.) ⇒ Archimedean Principle (AP)

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Proof. I will show that the contrapositive is true; that is, if ##\mathbb{F}## does not have the AP, it does not satisfy l.u.b.

By assumption, ##\mathbb{F}## has some element S such that for each ##n \in \mathbb{N}, n < S## (that is, it does not satisfy AP). It follows from the totally ordered field axioms that ##kS^{-1} < k/n## for all ##n, k \in \mathbb{N}##. The set ##\{S^{-1},2S^{-1},3S^{-1},4S^{-1},\cdots\}## is then bounded from above (setting ##n=k## shows that ##kS^{-1}## is always less than ##1##) but does not have an upper bound in ##\mathbb{F}##.

Suppose that it does have a least upper bound ##l \in \mathbb{F}##, and so for any two upper bounds ##l, l’ \in \mathbb{F}, l≤l’##. Then ##l## must hold to the inequality ##kS^{-1} < l## for all ##k## (otherwise, ##l## wouldn’t be an upper bound for ##S##). But then ##2kS^{-1}< l ## implies that ##kS^{-1}< l/2 < l##, and ##l/2## is a smaller upper bound for ##S##, contrary to ##l## being the least upper bound. This means that our supposition that ##S## has an upper bound is false, and the l.u.b. does not hold for ##\mathbb{F}##. □

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Definitions and Notes:

l.u.b. holds for ##\mathbb{F}## iff every non-empty set ##S \in \mathbb{F}## that is bounded from above has a least upper bound.

AP holds iff for any element ##s \in \mathbb{F}## there exists an ##n \in \mathbb{N}## such that ##s ≤ n.##

1 is the addition identify element for field addition, and multiplication by natural numbers is shorthand for repeated field addition. E.g., ##2S^{-1} = S^{-1} + S^{-1}##, and ##kS^{-1} = S^{-1} + S^{-1} + ... S^{-1}##, k times. If a natural number ##n## is used as an element of ##\mathbb{F}##, it is assumed to be repeated addition of the identity element n times.

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