# FOM + The Set of Real Numbers

## The Real Numbers and the Axiom of Completeness

As mentioned in our textbook (but notated strangely), we will be working with lots of sets in this class.  In particular, we are interested in the sets $\varnothing \subset \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}.$

It is this final set — that of the real numbers — that we will focus much of our attention on.  This set can be characterized by the following properties.

0)  There exist functions $+ : \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ and $\cdot : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ and a relation $<$ on $\mathbb{R}$.

1. $(x+y)+ z = x + (y+z)$ ; $(x\cdot y)\cdot z = x\cdot(y\cdot z)$
2. $x+y = y+x$ ; $x\cdot y = y\cdot x$
3. $x\cdot (x+z) = (x\cdot y) + (x\dot z)$
4. $\, \exists\, ! \, 0 \in \mathbb{R}, 0 + x = x$
5.  $\, \forall x \in \mathbb{R}, \,\, \exists \, \, y \in \mathbb{R}, x+y=0$, and we write $y = -x$ (additive inverses exist)
6. $\exists \, \, ! 1 \in \mathbb{R}, x\dot 1 = x$ for all $x \in \mathbb{R}$, and $0 \neq 1$.
7. $\forall \, \, x \in \mathbb{R}$ with $x \neq 0, \, \, \exists \, \, ! \, y \in \mathbb{R}, x\cdot y = 1$ and we write $y = x^{-1} = 1/x$.
8. $x < y \Rightarrow x+z < y+z$
9. $x < y$ and $y < z \Rightarrow x < z$
10. $\forall \, \, x, y \in \mathbb{R}$ exactly one of the following is true: $x < y, y < x$ or $x = y$.  (law of trichotomy)
11. $x < y$ and $z > 0 \Rightarrow xz < yz.$

The set with the two operations of $+$ and $\cdot$ and with the order relation $<$ — which we can conveniently notate as $\left(\mathbb{R}, +, \cdot, <\right)$ — is called an ordered field when all 11 of these axioms hold.

There is one more axiom we need to add to our list above, and this axiom is a bit bizarre sounding.  It doesn’t use the operations $+$ or $\cdot$, and it sounds a bit confusing and/or arbitrary when one first encounters it.  It’s called The Axiom of Completeness, and here’s what it says:

12. (The Completeness Axiom) Every non-empty subset of real numbers that is bounded from above has a supremum.

Why is this axiom so important?  Why does it seem so different from the previous axioms?  Next semester we will revisit these basic features of the real numbers, but instead of asserting them to be true, we will actually examine a construction of the set $\mathbb{R}$ (assuming the existence of the rationals, $\mathbb{Q}$), and we will show that all twelve of these axioms hold.  At that point it will be worth reminding everyone of a deep theorem (one we will state without proof):

Thm.  There is only one ordered field that satisfies Axioms 1-12.

For our first crack at Analysis, we will use these 12 statements as given facts about $\mathbb{R}$, and we will develop a basic understanding as to why the Completeness Axiom is so important.

The basic idea is this: the Completeness Axiom guarantees that every real number that ought to be on the number line is actually there.  Granted, this axiom is a rather round-about way of saying or implying this important fact, but once we work with it a little bit (and work with the notion of sequences of real numbers), this idea will become a bit clearer.  For the time being, though, we can understand how this axiom fails for the ordered field $\mathbb{Q}$ by considering the (classic) example of the set

$S \{x \in \mathbb{Q} : x^2 < 2\} \subset \mathbb{Q}$

It is not hard to argue that this set $S$ is non-empty (for example $0 \in S$) and bounded above.  For instance, I claim that the rational number $2 \in \mathbb{Q}$ is one upper bound for $S$.  How can we argue this point?  Well, suppose we had $x \in S$ with $x > 2$.  Then by using the Axioms we could conclude that $x^2 > 2^2 = 4$, but this would contradict the fact that $x \in S$.  Therefore, every $x \in S$ satisfies $x \leq 2$.

But now the real question: Does the set $S$ have a least upper bound?  More to the point: does the set $S$ have a rational least upper bound?  That is, is there a number in $\mathbb{Q}$ that deserves to be called the least upper bound of $S$?

It takes a little bit of work to show (nothing too bad), but, as you might suspect, if the set $S$ were to have a least upper bound $M$, then $M$ would need to satisfy

$M^2 = 2$

As proved in our textbook (and as proved in virtually every FOM class ever), any number $M$ that satisfies $M^2 = 2$ is not rational.  Therefore, this set $S \subset \mathbb{Q}$ has no (rational) least upper bound!

However, since $\sqrt{2} \in \mathbb{R}$, this set does have a least upper bound as a subset of the reals.  Indeed, this can give us a way to define $\sqrt{2}$ — it is the (guaranteed to exist) least upper bound of the non-empty, bounded subset

$S = \{x : x^2 < 2\}$

Theorems 0.20 through 0.23 (on pages 23-25) in our textbook establish some important consequences of the Axiom of Completeness, but the previous 11 axioms can be used to establish lots of important facts, too; many of these results are collected in Theorem 0.19 (page 22).

## FOM Background

To write clear proofs in this class, we will need to use a lot of FOM knowledge.  In particular, we will need to understand how to construct

1.  Direct Proofs
2. Proofs of Contrapositives