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+.