# Apostol's Mathematical Analysis 1.6

1. Apr 1, 2009

### cordyceps

1. The problem statement, all variables and given/known data
Prove that every nonempty set of positive integers contains a smallest member. This is called the well-ordering principle.

3. 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.

2. Apr 2, 2009

### HallsofIvy

Staff Emeritus
What properties of positive integers do you have to work with?

3. Apr 4, 2009

### cordyceps

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.

4. Apr 4, 2009

### cordyceps

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.

5. Apr 4, 2009

### cordyceps

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.