How is my proof of the Archimedean property?

[Jamin2112]

Homework Statement

Prove the Archimedean property

Homework Equations

Know what a least upper bound is

The Attempt at a Solution

Assume that if a and b are positive real numbers, na≤b for all natural numbers n. Then the set S of all numbers na, where n is a natural number, has b as its least upper bound.

Let n' be a natural number such that b-∂ < ∂n' ≤ b. Then b < ∂(n'+1). Since n' is a natural number, n'+1 is a natural number, and so ∂(n'+1) is an element of S. But since b < ∂(n'+1), S cannot have b as its least upper bound, and we have a contradiction.

 

[ystael]

You have the right basic idea, but several details and phrasings are wrong.

Assume that if a and b are positive real numbers, na≤b for all natural numbers n.

This hypothesis has the quantifier in the wrong place; what you said is "suppose that, for all positive real numbers [tex]a[/tex] and [tex]b[/tex] and all natural numbers [tex]n[/tex], we have [tex]na \leq b[/tex]". That's not the negation of the archimedean property. The correct hypothesis is "suppose that [tex]a[/tex] and [tex]b[/tex] are [specific] positive real numbers such that, for all natural numbers [tex]n[/tex], we have [tex]na \leq b[/tex]."

Then the set S of all numbers na, where n is a natural number, has b as its least upper bound.

This is a false deduction. You cannot conclude that [tex]\sup S = b[/tex], only that [tex]\sup S \leq b[/tex], because [tex]b[/tex] is an upper bound for [tex]S[/tex].

Let n' be a natural number such that b-∂ < ∂n' ≤ b. Then b < ∂(n'+1). Since n' is a natural number, n'+1 is a natural number, and so ∂(n'+1) is an element of S. But since b < ∂(n'+1), S cannot have b as its least upper bound, and we have a contradiction.

I'm not sure where the [tex]\partial[/tex] symbol came from. But if you substitute [tex]\sup S[/tex] for [tex]b[/tex], and [tex]a[/tex] for [tex]\partial[/tex], the rest of the argument is correct.

 

[Jamin2112]

Hmmmm ... Here's how my handout defines "upper bound" and "least upper bound":

Suppose that F is an ordered field and S is a nonempty subset of F. An upper bound for S is any element M of F so that x≤M for all x in S. A least upper bound for S is any element L of F which is an upper bound for S and which also has the property that every a<L is not and upper bound for S: L is the smallest upper bound for S.



So if I start with the assumption (rewritten the correct way)

"Suppose for all integers n, an≤b for some integers a and b."

Looking at the definitions, I see that any M≥b is an upper bound for the set S={na for all integers n and some positive real number a}. The least upper bound, of course, would be b.

Right?

 

[ystael]

So if I start with the assumption (rewritten the correct way)

"Suppose for all integers n, an≤b for some integers a and b."

This is also not right. Perhaps it's easier to see if you write it in symbols. Abbreviate the set of positive reals as [tex]\mathbb{R}_{>0}[/tex].

Your first attempt at the hypothesis was: [tex]\forall a \in \mathbb{R}_{>0}.\, \forall b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}.\, an \leq b[/tex].

Your second attempt translates to: [tex]\forall n \in \mathbb{N}. \, \exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, an \leq b[/tex]. You can tell this is not the right hypothesis, because it's true: for any natural number [tex]n > 0[/tex], one may choose [tex]a = \frac1{2n}, b = 1[/tex], and then [tex]an = \frac12 < 1 = b[/tex]. (For [tex]n = 0[/tex] just choose [tex] a = b = 1[/tex].)

The correct hypothesis is the negation of the archimedean property. The archimedean property is [tex]\forall a \in \mathbb{R}_{>0}. \, \forall b \in \mathbb{R}_{>0}. \, \exists n \in \mathbb{N}. \, an > b[/tex], so its negation is [tex]\exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}. \, an \leq b[/tex]. If you don't know how to negate statements with strings of quantifiers like this, you need to learn before you go much further in analysis.

Looking at the definitions, I see that any M≥b is an upper bound for the set S={na for all integers n and some positive real number a}. The least upper bound, of course, would be b.

Right?

No. Just because [tex]b[/tex] is the smallest upper bound you see in front of your face does not mean it is actually the least upper bound. You must prove that any other upper bound is at least [tex]b[/tex], and you can't do that from the hypotheses at hand.

 

[Jamin2112]

No. Just because [tex]b[/tex] is the smallest upper bound you see in front of your face does not mean it is actually the least upper bound. You must prove that any other upper bound is at least [tex]b[/tex], and you can't do that from the hypotheses at hand.

Hmmm ... but since there is an upper bound b, there must a least upper bound M≤b.