Least Upper Bound Property ⇒ Archimedean Principle

[Someone2841]

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.

 

[Someone2841]

This post got moved to abstract algebra (I was thinking of $\mathbb{F}$ as a generalization of the real numbers), and so I realize now that I could have just proven that any linearly-ordered group $\mathbb{G}$ that is not Archimedean does not have the least upper bound property. It can be done similarly but more concisely:

If $G$ is non-Archimedean linearly-ordered group, then there exists some $g, h \in G$ such that $g^n < h$ for all $n \in \mathbb{N}$. This means that the set $H=\{g^n : n \in \mathbb{N}\}$ is both non-empty and bounded by $h$.

Now suppose $H$ has a least upperbound $l \in G$. As a upper bound, $g^{n+1}<l$ for all $n$ (since $n$ is arbitrary). But this means that $g^n < g^{-1}l < l$, and $g^{-1}l$ is a lower least upper bound for $H$, which is a contradiction. Therefore, if $G$ is a non-Archimedean linearly-ordered group, it does not satisfy the least upper bound property. □