MATH 3150 Class notes 01-28-2025

notereal-analysis

As we’ve covered, $\mathbb{Q}$ satisfies the axioms of closure and associativity under addition, multiplication, and the distribution law so it is a field. Furthermore, $\mathbb{Q}$ satisfies the axiom of order, so it is an ordered field.

Goal: extend $\mathbb{Q}$ in a unique way to $\mathbb{R}$ by adding an additional axiom.

The Completeness Axiom

This is the axiom that differentiates the real numbers from the rational numbers.

Necessary Definitions

• Defined maximum
• Defined minimum
• Defined supremum
• Defined infimum

Writing Proofs with Supremum and Infimum

Example

Suppose $S=(-\infty ,1)=\{x\in \mathbb{R}\mid x<1\}$.

Claim: $M=1$ is the supremum of $S$.

Proof

We know that $M=1$ is an upper bound of $S$ because for all $x\in S$, $x<1$.

Suppose that $M^{\prime }$ is another upper bound of $S$. We want to show that $M\le M^{\prime }$.

Assume for the sake of contradiction that $M^{\prime }<M$, so $M^{\prime }<1$. Hence by the definition of $S$, $M^{\prime }\in S$, so $M^{\prime }$ is the maximum of $S$. However, we know that $S$ does not have a maximum (suppose $x\in S$ is the maximum of $S$, then we have that $(x+1)/2\in S$, so $x$ is not the maximum of $S$). Thus a contradiction has been reached and $M\le M^{\prime }$.

Defined The Completeness Axiom

Why does the completeness axiom not hold in the rational numbers? Suppose we had a set $Q\subset \mathbb{Q}$, defined by $Q=\{x\in \mathbb{Q}\mid x<\sqrt{2}\}$. We can not identify a supremum for $Q$ because for any $a\in \mathbb{Q}$ that we suppose is the supremum of $Q$, we could find $a^{\prime }\in \mathbb{Q}$ such that $\sqrt{2}<a^{\prime }<a$, so $a$ could not be the supremum of $Q$.

Defined Archimedean Property