Why does my topology textbook (Munkres) define positive integers as the intersection of all inductive subsets of the reals?

Question

This is how the topology textbook I'm reading (Munkres) defines integers:

A subset of the real numbers is "inductive" if it contains 1 and $1+x$ for all $x$ in the subset. The intersection of all inductive subsets of the reals is the set of positive integers.

Why take this route involving the intersection of so many sets? I could define the positive integers given reals as $1$ along with any sum of positive integers and get the same set much more easily.

Answer 1

You appear to be asking why we can't simply define $\mathbb N$ as an inductive set, i.e. a subset of $\mathbb R$ satisfying the following two axioms:

1. $1\in\mathbb N$
2. $(\forall x)(x\in\mathbb N\to x+1\in\mathbb N)$

The issue is that there are many sets which satisfy these two axioms. While the usual natural numbers are indeed an inductive set, so is:

• The set of positive real numbers.
• The set of positive real numbers unequal to $1/2$.
• The real numbers themselves!

Therefore, we need to add a third axiom to make our definition of $\mathbb N$ workable. One possibility, which is very similar in spirit to Munkres' definition, is the following:

3. $\mathbb N$ is the "smallest" set satisfying (1) and (2), i.e. if $X\subseteq\mathbb R$ is inductive, then $\mathbb N\subseteq X$.

However, we might find it difficult to rigorously prove that there is a subset $\mathbb N$ of $\mathbb R$ which satisfies these three axioms. To get around this issue, we could instead use Munkres' definition of $\mathbb N$ as the intersection of all inductive sets, before proving that $1\in \mathbb N$ and $(\forall x)(x\in\mathbb N\to x+1\in\mathbb N)$. Then, it immediately follows from the definition of intersections that $\mathbb N$ satisfies (3).

This is in fact exercise 25 of chapter 2 of Spivak's Calculus (4th edition), page 34.