Show that if \(\mathcal{A}\) is a collection of inductive sets, then the intersection of the elements of \(\mathcal{A}\) is an inductive set.
Prove the basic properties (1) and (2) of \(\mathbb{Z}_{+}\) .
Solution
Let \(\mathcal A\) be a collection of inductive subsets of \(\mathbb R\), and put \[ I=\bigcap_{A\in\mathcal A}A. \] Every \(A\in\mathcal A\) contains \(1\), so \(1\in I\). If \(x\in I\), then \(x\in A\) for every \(A\in\mathcal A\). Since each \(A\) is inductive, \(x+1\in A\) for every \(A\), hence \(x+1\in I\). Thus \(I\) is inductive.
By definition, \(\mathbb Z_+\) is the intersection of all inductive subsets of \(\mathbb R\). By part (a), \(\mathbb Z_+\) is itself inductive. This is property (1).
For property (2), let \(A\subset\mathbb Z_+\) be an inductive set of positive integers. Since \(A\) is an inductive subset of \(\mathbb R\), and \(\mathbb Z_+\) is the intersection of all such subsets, \[ \mathbb Z_+\subset A. \] But by hypothesis \(A\subset\mathbb Z_+\). Hence \[ A=\mathbb Z_+, \] which is the principle of induction.