# Proof of Well Ordering on N from Completeness & Archimedean Property

• ╔(σ_σ)╝
In summary: The well ordering principle is equivalent to complete and mathematical induction. So it would seem that I can "prove" the well ordering principle with completeness and the archimedean principle.
╔(σ_σ)╝

## Homework Statement

I am just curious ?

I have a feeling that completeness or the archimedean property relies on well ordering but I am not entirely sure.

However, completeness funishes a supremum or infimum for any subset of R that is bounded above or below, respectively.

## The Attempt at a Solution

$$N \subset R$$

So if S is any non empty subset of N then, S is a subset of R.

If S bounded below, it has a infima in R.

By the archimedean principle we can find can an integer that is greater or equal to the infima which would be in S.Is there something that prevents me from doing this ? Like completeness relying on well ordering of N.

I am not sure what you are asking. "Well ordering" of the natural numbersw does NOT follow from the completeness of the real numbers nor the Archimedean property because the natural numbers can be defined independently of the real numbers- in fact, normally, one defines the natural numbers first, then the integers from them, then the rational numbers from them, and the real numbers from the rational numbers. The "well ordering" property- every set of natural numbers has a smallest member follows from the "induction" axiom.

I notice, on re-reading, that, while you titled this "Can the well ordering on N be proven from completeness and archimedean property?", in your post you ask the opposite question- do "completeness" and the "archimedian property" depend on well ordering. The answer is no because we can construct other sets of numbers, for example the set of rational numbers, that have all the properties of the real numbers except completeness.

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HallsofIvy said:
I am not sure what you are asking. "Well ordering" of the natural numbersw does NOT follow from the completeness of the real numbers nor the Archimedean property because the can be defined independently of the real numbers- in fact, normally, one defines the natural numbers first, then the integers from them, then the rational numbers from them, and the real numbers from the rational numbers. The "well ordering" property- every set of natural numbers has a smallest member follows from the "induction" axiom.

I notice, on re-reading, that, while you titled this "Can the well ordering on N be proven from completeness and archimedean property?", in your post you ask the opposite question- do "completeness" and the "archmedian property" depend on well ordering. The answer is no because we can construct other sets of numbers, for example the set of rational numbers, that have all the properties of the real numbers except completeness.
My question should have been "does the construction of R depend on well ordering ".

Basically to ...

1)Prove the well ordering principle by using completeness and the archimedean principle ( That is what the sketch of a proof in my original post was)
2) If the archimedean property depened on the truth of well ordering then I obviously can not use it in the proof.

From what I understand the well ordering is equivalent to complete and mathematical induction. So it would seem that I can "prove" the well ordering principle with completeness and the archimedean principle.

My question about completeness depending on well ordering was retarded; I wasn't thinking when I aksed that question.Is my thinking correct? I sketched the idea of my "proof " in my original post. Is it valid ?

Wait, I don't even need the archimedean property.

Since, $$S \subset N \Rightarrow S \subset R$$

By completeness, inf(S) exist since S is bounded below eg 1 is a lower bound of S.

The fact that inf(S) is an integer is clear.

In fact, inf(S) is in S.

Isn't that the well ordering principle ?

╔(σ_σ)╝ said:
Wait, I don't even need the archimedean property.

Since, $$S \subset N \Rightarrow S \subset R$$

By completeness, inf(S) exist since S is bounded below eg 1 is a lower bound of S.

The fact that inf(S) is an integer is clear.

In fact, inf(S) is in S.

Isn't that the well ordering principle ?
You got me confused! Why is it "clear" that inf(S) is an integer? Assuming that it is, why does it follow, without using the "well ordering principal" that inf(S) is in S?

Why is it "clear" that inf(S) is an integer?
I am working on a proof to show that the least upper bound of any set of integers is an interger.

I showed that if inf(S) was not an integer then I looked at it's decimal expansion and found that I could select epsilon small enough that I could find a better upper bound which means inf(S) is an integer.

But then I discovered that it depends on how the integers are defined.
0.99999999999999... =1
Assuming that it is, why does it follow, without using the "well ordering principal" that inf(S) is in S?
If not then there is some $$s_{0}$$
$$s_{0} < inf(S) +1 \Rightarrow s_{0}-1<inf(S)\leq s_{0}$$
Inf(S) must equal $$s_{0}$$ since there is no integer between $$s_{0}$$ and
$$s_{0}-1$$

HallsofIvy said:
You got me confused! Why is it "clear" that inf(S) is an integer? Assuming that it is, why does it follow, without using the "well ordering principal" that inf(S) is in S?

## 1. What is the proof of well ordering on N from completeness and Archimedean property?

The proof of well ordering on N from completeness and Archimedean property is a mathematical proof that shows the natural numbers (N) are well ordered, meaning that every non-empty subset of N has a least element. This proof relies on two key properties of the real numbers: completeness, which states that every non-empty subset of real numbers has a least upper bound, and the Archimedean property, which states that for any positive real number, there exists a natural number that is larger than it.

## 2. Why is the proof of well ordering on N important?

The proof of well ordering on N is important because it is the foundation for many important mathematical concepts and theorems. It allows us to make statements about all natural numbers, rather than just specific ones, and it is crucial in the development of mathematical induction and other proof techniques.

## 3. How does the proof of well ordering on N relate to the principle of mathematical induction?

The proof of well ordering on N is closely related to the principle of mathematical induction. In fact, the principle of mathematical induction is a direct consequence of the well ordering of N. This is because the principle of mathematical induction relies on the fact that every non-empty subset of N has a least element, which is guaranteed by the well ordering of N.

## 4. Can the proof of well ordering on N be extended to other sets of numbers?

Yes, the proof of well ordering on N can be extended to other sets of numbers, such as the integers (Z) and the rational numbers (Q). This is because both Z and Q have the same completeness and Archimedean properties as the real numbers, which are necessary for the proof of well ordering on N.

## 5. Are there any real numbers that do not have a well ordering?

No, all real numbers have a well ordering. This is a fundamental property of the real numbers and is a consequence of the completeness property. This means that for any non-empty subset of real numbers, there will always be a least element, regardless of how large or small the subset is.

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