Skip to content

Mathematics · Ch 8 — Principle of Mathematical Induction

Natural Numbers as the Least Inductive Subset of Real Numbers

1

Natural Numbers as the Least Inductive Subset of Real Numbers

Before stating the principle of mathematical induction formally, it helps to ask a more basic question: what actually characterises the set of natural numbers N={1,2,3,… }N = \{1, 2, 3, \dots\} inside the much larger set of real numbers R\mathbb{R}?

A subset that starts at 11 and keeps stepping forward. Call a subset S⊆RS \subseteq \mathbb{R} inductive if it satisfies two closure conditions: (i) 1∈S1 \in S, and (ii) whenever a real number kk belongs to SS, the next number k+1k+1 also belongs to SS. Many subsets of R\mathbb{R} are inductive in this sense -- for instance R\mathbb{R} itself is inductive, and so is the interval [1,∞)[1, \infty). But among all such inductive subsets of R\mathbb{R}, the set of natural numbers NN is the smallest one: NN is precisely the intersection of every inductive subset of R\mathbb{R}, or equivalently, NN contains 11, contains k+1k+1 whenever it contains kk, and contains nothing else.

Why this matters for proofs. This least-inductive-subset description of NN is exactly what makes a two-step argument enough to establish a statement P(n)P(n) for every natural number nn. If a set S={n∈N:P(n) is true}S = \{ n \in N : P(n) \text{ is true} \} can be shown to satisfy the two inductive closure conditions -- 1∈S1 \in S (i.e. P(1)P(1) holds), and k∈S⇒(k+1)∈Sk \in S \Rightarrow (k+1) \in S (i.e. P(k)P(k) holds forces P(k+1)P(k+1) to hold) -- then SS is itself an inductive subset of R\mathbb{R} contained in NN. But NN is the smallest inductive subset, so SS cannot be a proper subset of NN; it must equal all of NN. In other words, P(n)P(n) is true for every n∈Nn \in N.

A step-ladder picture. It is often useful to picture the natural numbers as rungs of an infinitely tall ladder. If you know you can stand on rung 11, and you know that whenever you are standing on any rung kk you can always climb to rung k+1k+1, then you can conclude you can reach every rung, however high. This intuition -- a starting foothold, plus a reliable step that always carries you one rung further -- is precisely the content of the principle stated formally in the next section.