Apostol's Mathematical Analysis 1.6

cordyceps
Messages
50
Reaction score
0

Homework Statement


Prove that every nonempty set of positive integers contains a smallest member. This is called the well-ordering principle.


The Attempt at a Solution


I'm just starting out with analysis, so I'm not too sure about the format of proofs. Here goes:

Proof. First suppose the set S of positive integers is finite. Then assume S contains no smallest member. It follows that for every x in S, there exists an infinite number of members of S less than x. This contradicts our assumption that S is finite. Thus every finite set of integers contains a smallest member.

Any help/critique is appreciated.
 
Physics news on Phys.org
What properties of positive integers do you have to work with?
 
Sorry for the late reply. I have the definition of an inductive set and the definition of a positive integer: a real number which belongs to every inductive set.
 
Oh, also some of the field axioms for real numbers:
1. Exactly one of the relations x = y, x < y, x > y holds.
2. If x > y and y > z, then x > z.
 
So, for the non-empty set S of positive integers, one of the relations x < y or x > y holds between any two members. This means that each member in S is greater than or less than each of the other members. Also, if some member x of S is less than some member y, and y is less than some member z, then x is less than z. It follows that there is a member less than every other member in S.

Is this clear enough? Intuitively, I see why the theorem is true, but I don't think I present my thoughts very rigorously.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top