Least Upper Bound Property ⇒ Archimedean Principle

• Someone2841
In summary, the proof shows that if a totally ordered field ##\mathbb{F}## does not satisfy the Archimedean Principle, it also does not satisfy the Least Upper Bound Property. This is proven by showing that a set bounded from above in ##\mathbb{F}## does not have a least upper bound, leading to a contradiction. The definitions of l.u.b. and AP are also provided, as well as a note on using natural numbers in ##\mathbb{F}## as repeated field addition. The proof can also be generalized to any non-Archimedean linearly-ordered group.

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.

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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. □

What is the Least Upper Bound Property?

The Least Upper Bound Property, also known as the Completeness Property, states that every non-empty set of real numbers that is bounded above has a least upper bound (supremum).

What is the Archimedean Principle?

The Archimedean Principle states that for any two positive real numbers a and b, there exists a positive integer n such that na > b.

How are the Least Upper Bound Property and Archimedean Principle related?

The Least Upper Bound Property and Archimedean Principle are both fundamental properties of the real number system. The Archimedean Principle is a consequence of the Least Upper Bound Property, as it can be proven using the property's definition.

What are some examples of sets that satisfy the Least Upper Bound Property?

Some examples of sets that satisfy the Least Upper Bound Property are closed intervals [a, b], where a and b are real numbers, and the set of all positive real numbers.

Why are the Least Upper Bound Property and Archimedean Principle important in mathematics?

The Least Upper Bound Property and Archimedean Principle are important because they are fundamental properties that allow us to make precise statements and proofs in mathematics, particularly in calculus and real analysis. These properties also help us understand the behavior of real numbers and their relationships with other mathematical concepts.