# Set Theory Proof. Inductive sets.

1. Oct 4, 2012

### jmjlt88

Claim: If A is an inductive set of postive integers, then A is Z+.

I tried to prove this two different ways for the fun of it. I would like to get some feedback concerning the correctness of both. Thank you. :tongue:

Proof: By definition, Z+ is the intersection of all inductive subsets of ℝ. Since A is an inductive subset of ℝ, it follows that Z+ is a subset of A. Next, take any a ε A. Then since A is a set of positive integers, a ε Z+. This gives us our desired result.

Proof: First, let us note that since A is inductive, 1 ε A, and if n ε A, then n+1 ε A. Now, we also know A is a set of positive integers. To show that A is indeed all of Z+, let us assume the contrary. Let n be the smallest positive integers not in A. Then, n-1 is in A, which implies (n-1)+1 is in A. Hence, n is in A and we have arrived at a contradiction. Thus, A must be all of Z+.

2. Oct 4, 2012

### micromass

Staff Emeritus
How do you know such a smallest positive integer exists??

3. Oct 4, 2012

### jmjlt88

Perhaps a misuse of the well-ordering property?

4. Oct 4, 2012

### micromass

Staff Emeritus
Sure, the well-ordering property applies. But before you can apply the well-ordering theorem, you need to prove that it holds true first. But I think that the proof of the well-ordering property uses the claim in the OP (especially if you defined $\mathbb{Z}^+$ as the intersection of inductive sets). If you already proved the well-ordering property, then your proof is correct.

By the way, the first proof in the OP seems correct no matter what.

5. Oct 4, 2012

### jmjlt88

Thanks! =)

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