Real Analysis: Lecture Notes

1 The Real Numbers

1.1 Bounds and the Completeness Axiom

To rigorously study calculus, we must understand the fundamental properties that distinguish the real numbers $ℝ$ from the rational numbers $ℚ$ . The most crucial of these is the Completeness Axiom.

Definition 1.1. Let $𝑆 \subset ℝ$ be a non-empty set. A number $𝑀 \in ℝ$ is called an upper bound for $𝑆$ if $𝑥 \leq 𝑀$ for all $𝑥 \in 𝑆$ . If $𝑆$ has an upper bound, we say $𝑆$ is bounded above.

Definition 1.2. Let $𝑆$ be bounded above. The supremum (or least upper bound) of $𝑆$ , denoted $sup 𝑆$ , is a number $𝑠 \in ℝ$ such that:

$𝑠$ is an upper bound for $𝑆$ .
If $𝑢$ is any upper bound for $𝑆$ , then $𝑠 \leq 𝑢$ .

Theorem 1.3. (The Completeness Axiom) Every non-empty subset of $ℝ$ that is bounded above has a supremum in $ℝ$ .

1.2 The Archimedean Property and Density

The Completeness Axiom allows us to prove several essential properties of the real numbers.

Theorem 1.4. (Archimedean Property) For every real number $𝑥 \in ℝ$ , there exists a natural number $𝑛 \in ℕ$ such that $𝑛 > 𝑥$ .

Proof. Assume for the sake of contradiction that this is false. Then there exists an $𝑥 \in ℝ$ such that $𝑛 \leq 𝑥$ for all $𝑛 \in ℕ$ . This implies that $ℕ$ is bounded above by $𝑥$ .

By Theorem 1.3, $ℕ$ must have a supremum in $ℝ$ . Let $𝑠 = sup ℕ$ . Since $𝑠$ is the least upper bound, $𝑠 - 1$ cannot be an upper bound for $ℕ$ . Therefore, there exists some $𝑚 \in ℕ$ such that $𝑚 > 𝑠 - 1$ .

Rearranging this gives $𝑚 + 1 > 𝑠$ . Since $𝑚 \in ℕ$ , closure under addition implies $𝑚 + 1 \in ℕ$ . However, this means we have found a natural number strictly greater than our supremum $𝑠$ , which contradicts the definition of an upper bound. Thus, our initial assumption must be false. $□$

Lemma 1.5. (Density of the Rationals) If $𝑥, 𝑦 \in ℝ$ with $𝑥 < 𝑦$ , then there exists a rational number $𝑟 \in ℚ$ such that $𝑥 < 𝑟 < 𝑦$ .

Proof. Since $𝑥 < 𝑦$ , we have $𝑦 - 𝑥 > 0$ . By Theorem 1.4, there exists $𝑛 \in ℕ$ such that $𝑛 (𝑦 - 𝑥) > 1$ , or $𝑛 𝑦 - 𝑛 𝑥 > 1$ .

Since the distance between $𝑛 𝑥$ and $𝑛 𝑦$ is strictly greater than $1$ , there must exist an integer $𝑚 \in ℤ$ strictly between them:

𝑛 𝑥 < 𝑚 < 𝑛 𝑦

Dividing by $𝑛$ , we obtain:

𝑥 < \frac{𝑚}{𝑛} < 𝑦

Setting $𝑟 = \frac{𝑚}{𝑛}$ , we have found our rational number, completing the proof. $□$