Set Theory Proof. Inductive sets.

[jmjlt88]

Claim: If A is an inductive set of positive 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.

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₊.

 

[micromass]

Let n be the smallest positive integers not in A.

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

 

[jmjlt88]

Perhaps a misuse of the well-ordering property?

 

[micromass]

Perhaps a misuse of the well-ordering property?

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.

 

[jmjlt88]

Thanks! =)