The Timpson Lyceum · Real Analysis (2020)

Lecture 2 (Real Analysis 2020) part 3

Overview

This part builds the natural numbers up from the notion of an inductive set, then uses that foundation to justify the Principle of Mathematical Induction as a theorem rather than an article of faith. With induction in hand we climb the number systems — integers, rationals — and discover that even though the rationals satisfy every field and positivity axiom, they are not enough: there is no rational number whose square is $2$ . Proving $\sqrt{2}$ is irrational motivates the final ingredient of the real numbers, the Completeness Axiom, expressed through least upper bounds (suprema). The part closes by defining the square-root symbol via a uniqueness result and showing that completeness for upper bounds automatically delivers greatest lower bounds (infima).

Key Concepts

Inductive set. A set $S \subseteq \mathbb{R}$ is inductive if $1 \in S$ and whenever $x \in S$ then $x+1 \in S$ .

The natural numbers. $\mathbb{N}$ is defined as the intersection of all inductive sets. Consequently:

$\mathbb{N}$ is a subset of every inductive set.
$\mathbb{N}$ is itself inductive: $1$ lies in every inductive set so $1 \in \mathbb{N}$ ; and if $x \in \mathbb{N}$ then $x$ lies in every inductive set, so $x+1$ does too, hence $x+1 \in \mathbb{N}$ .
If $A$ is an inductive subset of $\mathbb{N}$ , then $A = \mathbb{N}$ (since $\mathbb{N} \subseteq A$ from inductiveness and $A \subseteq \mathbb{N}$ by assumption). This double-containment fact is the engine behind every proof by induction.

Principle of Mathematical Induction. Let $S(n)$ be a statement about each natural number $n$ . If

(base case) $S(1)$ is true, and
(inductive step) $S(k) \Rightarrow S(k+1)$ for every $k \in \mathbb{N}$ ,

then $S(n)$ is true for all $n \in \mathbb{N}$ . The proof: form $A = \{\, n \in \mathbb{N} : S(n) \text{ is true} \,\}$ . The base case gives $1 \in A$ ; the inductive step gives $k \in A \Rightarrow k+1 \in A$ . So $A$ is an inductive subset of $\mathbb{N}$ , hence $A = \mathbb{N}$ . Induction only reaches the naturals (starting at $1$ and stepping by $+1$ never lands on $\tfrac12$ , say), which is exactly why the conclusion is restricted to $\mathbb{N}$ .

The number systems.

$\mathbb{Z}$ (integers) = the naturals, their negatives, and $0$ .
$\mathbb{Q}$ (rationals) $= \left\{\, \dfrac{m}{n} : m,n \in \mathbb{Z},\ n \neq 0 \,\right\}$ .
$\mathbb{N}$ fails the field axioms (no additive inverses, no $0$ ); $\mathbb{Z}$ fails them (no multiplicative inverses); $\mathbb{Q}$ is the first system satisfying all field and positivity axioms — yet still not the reals.

Divisibility, even, odd. For integers $a,b$ : $a \mid b$ means there exists $c \in \mathbb{Z}$ with $ac = b$ . An integer is even if it equals $2x$ for some $x \in \mathbb{Z}$ (equivalently $2 \mid a$ ); odd if it equals $2x+1$ for some $x \in \mathbb{Z}$ . Basic results proved by these definitions: a product of two evens is even; a product of two odds is odd; and (by induction) every integer is even or odd.

Why the rationals are not enough. The polynomial $x^2 - 2$ is negative at $x=0$ and positive at $x=2$ , so it "should" cross zero — but no rational number does so. Hence analysis needs more than the field/positivity axioms.

Bounded sets and the Completeness Axiom. A nonempty set $S \subseteq \mathbb{R}$ is bounded above if there is $c \in \mathbb{R}$ with $x \le c$ for all $x \in S$ (such a $c$ is an upper bound). The Completeness Axiom: every nonempty set that is bounded above has a least upper bound. This least upper bound is the supremum $\sup S$ . Symmetrically, a set is bounded below if some $c \le x$ for all $x \in S$ , and its greatest lower bound is the infimum $\inf S$ . For example $\{x : x^2 < 2\}$ is bounded above with supremum $\sqrt{2}$ — completeness is what conjures irrational numbers into existence.

Uniqueness of positive square roots. For each $c > 0$ there is exactly one $x > 0$ with $x^2 = c$ ; this unique positive solution defines $\sqrt{c}$ . Existence comes from completeness; uniqueness is proved below.

Infima come for free. Completeness is stated only for upper bounds, but reflecting a set through $0$ converts a lower bound into an upper bound, so every nonempty set bounded below has an infimum — no second axiom required.

Worked Example

Claim. $\displaystyle \sum_{j=1}^{n} j = \frac{n(n+1)}{2}$ for every $n \in \mathbb{N}$ (proof by induction).

Let $A = \left\{\, n \in \mathbb{N} : \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \,\right\}$ . Showing $A$ is inductive forces $A = \mathbb{N}$ .

\begin{aligned} \textbf{Base case }(n=1):\quad \sum_{j=1}^{1} j &= 1 = \frac{1\cdot 2}{2} = \frac{1(1+1)}{2}, \quad\text{so } 1 \in A. \\ \textbf{Inductive step: assume } \sum_{j=1}^{k} j &= \frac{k(k+1)}{2} \quad (k \in A). \\ \sum_{j=1}^{k+1} j &= \left(\sum_{j=1}^{k} j\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) \\ &= \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2} = \frac{(k+1)\big((k+1)+1\big)}{2}. \end{aligned}

So $k+1 \in A$ . Thus $A$ is inductive, $A = \mathbb{N}$ , and the formula holds for all $n$ .

A second, contradiction-style proof: $\sqrt{2} \notin \mathbb{Q}$ . Suppose for contradiction that $\sqrt{2} = \tfrac{m}{n}$ with $m,n \in \mathbb{Z}$ , $n \neq 0$ , where (reducing the fraction) at least one of $m, n$ is odd.

\begin{aligned} \frac{m^2}{n^2} &= 2 \;\Rightarrow\; m^2 = 2n^2 \;\Rightarrow\; m^2 \text{ even} \;\Rightarrow\; m \text{ even.} \\ m = 2k \;\Rightarrow\; m^2 = 4k^2 \;\Rightarrow\; 4k^2 = 2n^2 \;&\Rightarrow\; n^2 = 2k^2 \;\Rightarrow\; n^2 \text{ even} \;\Rightarrow\; n \text{ even.} \end{aligned}

Now $m$ and $n$ are both even, contradicting that at least one was odd. Hence no such fraction exists and $\sqrt{2} \notin \mathbb{Q}$ .

Summary

$\mathbb{N}$ is the intersection of all inductive sets; it is itself inductive, and its only inductive subset is $\mathbb{N}$ — the key fact behind induction.
The Principle of Mathematical Induction is a theorem: take $A = \{n : S(n)\}$ , show it is inductive, conclude $A = \mathbb{N}$ . Induction reaches exactly the naturals.
Climbing $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}$ , the rationals are the first system satisfying all field and positivity axioms, yet $x^2 = 2$ has no rational solution.
Using divisibility and the even/odd dichotomy (every integer is even or odd), a contradiction argument shows $\sqrt{2}$ is irrational.
The Completeness Axiom — every nonempty set bounded above has a least upper bound (supremum) — supplies the missing real numbers; the supremum of $\{x : x^2 < 2\}$ is $\sqrt{2}$ .
Each $c>0$ has a unique positive square root (defining $\sqrt{c}$ ), and reflecting through $0$ shows every set bounded below has an infimum, so no separate axiom for lower bounds is needed.