Proof of the Archimedean Property

Main Question or Discussion Point

I am reading Rudin's proof of this property, but I find one assertion he makes quite disagreeable to my understanding; I am hoping that someone could expound on this assertion. Here is the statement and proof of the archimedean property:

(a) If $x \in R$, $y \in R$, and $x > 0$, then there exists a positive integer $n$ such that $nx > y$.

(a) Let A be the set of all nx. If (a) were false, then y would be an upper bound of A. But then A has a least upper bound in R...

I understand the method of proof he is employing, a proof by contradiction, which leads to y being larger than every element A (hence, it is an upper bound). However, I do not understand how this immediately implies that A must have a least upper bound. Which theorem is he invoking to come to such a conclusion?

EDIT: Also, what is the quantification on x and y?

ShayanJ
Gold Member
I think its like this: If there is no n, such that $nx > y$, then y is greater than all the elements of A. But because the elements of A can be arbitrarily large, this means that $y=\infty$. But this is a contradiction because we chose y in the beginning and we could have chosen a finite real number.
And about your last question, the theorem can be written like below:
$\forall x,y \in \mathbb R \ \ \ (x>0 \rightarrow (\exists n\in \mathbb N \ \ \ (nx>y)))$

Stephen Tashi
I think its like this: If there is no n, such that nx>y nx > y , then y is greater than all the elements of A. But because the elements of A can be arbitrarily large, this means that y=∞ y=\infty . But this is a contradiction because we chose y in the beginning and we could have chosen a finite real number.
So, would the idea of $A$ having a least upper bound even be necessary for the proof?

ShayanJ
Gold Member
So, would the idea of $A$ having a least upper bound even be necessary for the proof?
Actually it isn't clear to me. That's why I proposed my own version of it. Maybe if you give the proof completely, I can tell!

EDIT:Well, I didn't use a least upper bound for A, so it doesn't seem necessary!

jbriggs444
Homework Helper
2019 Award
If A is the set of all real numbers of the form nx for some natural n and if y is an upper bound on A then by the least upper bound property, A must have a least upper bound. Call this bound B.

B is a real number. B minus x is a real number less than B. Since B is a least upper bound on A, B-x cannot be an upper bound on A. So there must be at least one member of A between B-x and B. That is, there is a real number of the form nx between B-x and B. Pick any such number and add x to it.

This new number is also of the form nx for some n. Thus it is a member of A and is greater than B. Contradiction.

Edit: Note that the proof offered by Shyan invokes infinity without definition and is, thus, invalid.

ShayanJ
Gold Member
Edit: Note that the proof offered by Shyan invokes infinity without definition and is, thus, invalid.
Actually I thought I have the idea right and people can fix the probable handwavy things in it. You mean it can't be fixed?

Normally you do not know that to each real number there is a bigger natural number when you prove this theorem.

jbriggs444
Homework Helper
2019 Award
Pulling that proof back up and looking at it...

I think its like this: If there is no n, such that $nx > y$, then y is greater than all the elements of A.
But because the elements of A can be arbitrarily large, this means that $y=\infty$.
But this is a contradiction because we chose y in the beginning and we could have chosen a finite real number.
$\infty$ is not a real number. So y cannot be equal to $\infty$. Further, there is no predicate "is finite" defined for the real numbers.

Edit: I missed a flaw. We didn't pick y in the beginning. The other team did. The assertion is that the relationship holds for all y. Even if there is such a thing as an infinite y, there still has to be an n large enough so that nx > y.

Last edited:
Pulling that proof back up and looking at it...

$\infty$ is not a real number. So y cannot be equal to $\infty$. Further, there is no predicate "is finite" defined for the real numbers.

Edit: I missed a flaw. We didn't pick y in the beginning. The other team did. The assertion is that the relationship holds for all y. Even if there is such a thing as an infinite y, there still has to be an n large enough so that nx > y.
Oops. Silly me. I thought we were on the same team. :D