Lecture 2 - Starting to Construct the Reals

Order!

We start with a definition on Order:

Order

A set is ordered if, given two elements $r, s$ from the set, exactly one of the following is true:

$r < s$
$r = s$
$r > s$
We also require transitivity, namely $\forall x, y, z$ in the set and given $x < y$ and $y < z$ , then $x < z$ .

Namely, if $S$ is a set, then an order on $S$ says that $<$ is a relation satisfying that $\forall x, y \in S$ exactly one of the above conditions holds. Transitivity is also a huge requirement.

Note

When we say $x > y$ we mean $y < x$ in the background. Further $x \leq y$ says $x = y$ or $x < y$ . Similar for the $\geq$ .

Compare the definition for bounds from the textbook as compared to lecture:

Bounds

A set $A \subseteq R$ is bounded above if there exists a number $b \in R$ such that $a \leq b$ for all $a \in A$ . The number $b$ is called an upper bound for $A$ .
Likewise, the set $A$ is bounded below if there exists a lower bound $l \in R$ where $l \leq a$ for every $a \in A$ .

Bounds

Let $S$ be an ordered set. Let $B \subseteq S$ . If $\forall b \in B$ we have $b \leq z$ , then $z$ is an upper bound.
Similarly, if $\forall b \in B$ we have $b \geq z$ then $z$ is a lower bound.

We'll say that $B$ is bounded above if $\exists$ an upper bound. Similarly $B$ is bounded below when $\exists$ a lower bound. Notice here we NEED to have the definition of Order in order to use these bounded ideas.

Suprema

Least Upper Bound (supremum)

Some $s \in R$ is the least upper bound for a set $A \subseteq R$ if it meets:

$s$ is an upper bound for $A$
if $b$ is any upper bound for $A$ then $s \leq b$
We denote the least upper bound as the supremum, or $sup (A)$ . Sometimes you'll set $s = lub (A)$ but we stick with the former.

Compare with the lecture one:

Least Upper Bound

Let $S$ be an ordered set. Let $B \subseteq S$ . Suppose $z \in S$ satisfies:

$z$ is an upper bound for $B$ .
If $y$ is any upper bound then $z \leq y$ .
Then $z$ is the least upper bound of $B$ .

Note

Notice the use of the word "the" when dictating least upper bound (lub). That's because these are unique, as we will later see.

We call $z$ the supremum of $B$ , denoted $sup (B)$ . Similarly for greatest lower bound is called an infimum, denoted $inf (B)$ .

Note

Here we are using a set $B$ , while the book uses $R$ . That's because we really don't need $R$ yet, as some of the examples we do will show.

Example

Let $S = Q$ here, and $B = {q \in Q : 0 \leq q < 1}$ . What are the lower bounds and upper bounds? Suprema? Infima?

Proof
Here are some examples of lower bounds:

- 2, - 1, - 0.1, - 0.2, . . ., 0

but here $0$ is the standout one, because it's the greatest lower bound. Thus $inf (B) = 0$ in this case (we haven't really proved it, but we're going by looks here. Just apply the definitions).

Examples of lower bounds are:

1, 2, 100, 1000000, . . .

But here $1$ is the smallest upper bound. Hence $1 = sup (B)$ .

Note

Notice that while $1 \notin B$ , we still have $sup (B) = 1$ . It's important that you don't have to have the element in the set to have a suprema or infima. Similarly, the bounds at all don't have to be necessarily in $B$ , just at least in $S$ .
☐

Example

Let $S = Q$ . Have $B = {\frac{1}{n} | n \in N}$ . What are the suprema and infima?

Proof
Here 1 is the smallest upper bound, so $sup (B) = 1$ . Similarly $0 = inf (B)$ .
☐

Example

Let $S = Q$ . Have $B = {q \in Q | q > 0}$ . What are the suprema and infima?

Proof
Here $inf (B) = 0$ , but there are no upper bounds, so clearly $∄ sup (B)$ here.
☐
But can you make a set where the suprema isn't in $S$ itself? Remember that $\sqrt{2} \notin Q$ so if we have $S = Q$ then all we need is a set $B$ where $sup (B) = \sqrt{2}$ . An example would be:

B = {q | q^{2} < 2}

clearly if $\sqrt{2} = sup (B)$ here, but $sup (B) = \sqrt{2} \notin S$ in this case, which is a contradiction. Hence $\sqrt{2}$ is not the suprema! There are other upper bounds like $5, 2, 1.5, 1.419, . . .$ but you can always get a smaller upper bound. We can get closer, but the condition that the suprema needs to be in $S = Q$ blocks us from having one in this case.

Theorem

If $S = Q$ and $B = {q | q^{2} < 2}$ here has no least upper bound.

Using our Field $F$ , we can add Order to get an Ordered Field for $F$ :

Ordered Field

An ordered field is a Field that has Order $<$ satisfying these additional properties:

$\forall x, y, z \in F$ , if $x < y$ then $x + z < y + z$ (additive preservation of $<$ )
$\forall x, y \in F$ where $x, y > 0$ , then $x y > 0$ (multiplicative preservation of $<$ )

Here notice that $Q$ already forms an Ordered Field. Heck even $R$ as we seem to know it right now also follow that. But $R$ has the gaps that $Q$ of these missing suprema. As such, we define $R$ to just have those suprema that $Q$ misses.

The Axiom of Completeness

Every nonempty set of real numbers that is bounded above has a least upper bound.

With this we can get:

We note that $R$ really is just an extension of $Q$ , filling in the irrational holes like $\sqrt{2}, \sqrt{3}, . . .$ This definition of $R$ really was chosen for this property of completeness, so even though the justification as to why this property is important will have to wait, you'll see why as we use this axiom a ton. Just know that by the 1870s, we had a pretty rigorous way to construct $R$ from $Q$ , which we'll get to see in 8.6.

An Initial Definition of $R$

First $R$ contains $Q$ with the same operations of addition, multiplication:

All elements $x \in R$ has their additive inverse and if $x \neq 0$ then the multiplicative inverse $x^{- 1}$ exists too.
$R$ is a Field, giving the properties of:
- Commutativity
- Associativity
- Distributive Property
$R$ also gets the Order from $Q$ to all of $R$ . So things like "If $a < b$ and $c > 0$ implies $a c < b c$ " are silently derived from this ordering.
Thus, $R$ is an Ordered Field, containing $Q$ as a subfield

This brings us to the idea of completeness. We need to say that $R$ doesn't have "any gaps" where the irrationals should go:

The Axiom of Completeness

Every nonempty set of real numbers that is bounded above has a least upper bound.

\exists R

$\exists$ an ordered field $R$ which contains $Q$ which satisfies the the axiom of completeness.

See Existence of the Reals for a proof/construction of these.

References

[[Abbott Real Analysis.pdf#page=27]]

(nothing todo!)

Order!

Suprema

An Initial Definition of R

References

An Initial Definition of $R$