# 1 - Dedekind completeness, Archimedean, and denseness of Q.
An ordered set $R$ is said to have the **least upper bound property**, if every nonempty subset $B\subset R$ that is bounded above has a least upper bound, namely $\sup B$ exists in $R$. The least upper bound property is also called **Dedekind completeness**.
The real numbers are axiomatically defined as follows:
> Real axioms.
>
> An *ordered field* $\mathbf{R}$ such that it has the *least upper bounded property (LUB)* is called the real numbers.
And indeed, we can construct such an ordered field that has the LUB property (via infinite decimal expansion, Dedekind cuts, Cauchy sequences, Eudoxus reals, etc. However the details can be cumbersome to check). Furthermore, any ordered field satisfying the real axiom are in fact the same up to field isomorphism. To this end, it makes sense to speak of *the* real numbers.
**Remark.** This axiomatic definition of the reals $\mathbf{R}$ captures the following aspects: Algebraic (field), geometric (ordered set), and analytic (least upper bound property).
**Remark.** In an ordered set $S$, if it satisfies the least upper bound property, then it also satisfies the greatest lower bound property, that every nonempty set $A$ bounded below has an infimum in $S$. $\blacktriangleright$ Take nonempty subset $A$ that is bounded below, so some lower bound $\ell$ exists. Denote $L$ to be the set of all lower bounds of $A$. Note $L$ is bounded above and not empty, so $a = \sup L$ exists in $S$. We claim $a = \inf A$. First, $a$ is a lower bound of $A$. If not, that there exists some $x \in A$ where $x < a$, then $x$ is not an upper bound of $L$. This means there is some $\ell \in L$ such that $x < \ell$. But $\ell$ is a lower bound of $A$, so $\ell \le x$, contradiction. Next $a$ is the greatest lower bound of $A$. If $\ell$ is another lower bound of $A$, then $\ell \in L$, so $\ell \le \sup L = a$ by definition. $\blacksquare$
## Some definitions.
An **ordering** $\le$ on a set $S$ is such that (1) $\le$ is *reflexive*, (2) $\le$ is *antisymmetric*, and (3) $\le$ is *transitive*. The ordering is a **total ordering**, or **linear ordering**, if for every two elements $a,b \in S$, we have $a\le b$ or $b\le a$, that all pairs are *comparable*.
An **ordered field** $F$ is a field equipped with a linear ordering $\le$ on $F$ such that (1) $a \le b$ implies $a + c \le b + c$, and (2) $0 \le a$ and $0 \le b$ implies $0 \le a b$. In other words, $\le$ respects sum and preserves positives. We can think of an ordered field (or any totally ordered set) as "numbers on a line".
For example, the usual rational numbers $\mathbf{Q}$ form an ordered field. An interesting example is the ordered field $\mathbf{Q}(x)$, all rational functions with indeterminate $x$, and we can prescribe a total ordering by $f < g$ if $f(x) < g (x)$ for all $x$ in some small enough interval $(0,\epsilon)$, some $\epsilon > 0$.
One can show that every ordered field contains an isomorphic copy of $\mathbf{Q}$, namely of [[supplemental/characteristic-of-a-field|characteristic]] zero.
///Why?///
Suppose $F$ has positive characteristic $p$, then $1+1+\cdots +1 =p\cdot 1= 0$. Suppose $0 < 1$, then by repeated adding $1$ we get $1 < 2$, $2 < 3$, ..., and $p-1 < 0$, which by transitivity we get $0 < 0$, a contradiction. Similarly if $1 < 0$, the same argument also shows $0 < 0$. Hence $F$ must have characteristic zero, and contains a copy of $\mathbf{Q}$. $\blacksquare$ ////
In an ordered set $(S,\le)$, an element $u\in S$ is an **upper bound** a subset $A \subset S$ if for every element $a\in A$, we have $a \le u$. A subset $A\subset S$ is **bounded above** if it has an upper bound $u \in S$.
An element $u \in S$ is said to be the **least upper bound** of some subset $A\subset S$ if (1) $u$ is an upper bound of $A$, and (2) for every upper bound $u'$ of $A$, we have $u \le u'$. Note if the least upper bound of some subset $A$ exists, then it is necessarily unique, which we can call it *the* least upper bound of $A$, and denote $u = \sup A$.
A useful principle in an ordered field $F$ is that if $u = \sup A$ for some subset $A\subset F$, then for any positive $\epsilon > 0$, $u-\epsilon$ is no longer an upper bound of $A$, and therefore there exists some $a\in A$ such that $a > u- \epsilon$.
Symmetrically we can define **lower bound**, **bounded below**, and **greatest lower bound** in an ordered set. Note vacuously, an empty subset is bounded above and below, and every element is an upper bound, as well as a lower bound, of the empty subset.
---
Not every nonempty subset in an ordered set has a least upper bound, however, even when it is bounded above. Indeed, the rationals $\mathbf{Q}$ suffers from this, we can show that the set $A = \{r\in \mathbf{Q} : r^2 < 2\}$ is a nonempty subset that is bounded above, but $\sup A$ does not exist in $\mathbf{Q}$.
///Why?///
Suppose to the contrary that $u = \sup A$ exists in $\mathbf{Q}$. We shall show $u^{2}$ cannot be smaller than 2, nor be greater than 2. So we must have $u^{2} =2$, which is a contradiction as there is [[supplemental/rational-roots-of-monic-integral-poly-are-integers|no rational whose square is 2.]]
Now, suppose $u^{2} < 2$. Consider $u + \frac{1}{N}$ for some positive $N$. Note that $$
\left( u + \frac{1}{N} \right)^{2} = u^{2} +\frac{2u}{N} +\frac{1}{N^{2}} < u^{2} + \frac{2u + 1}{N}.
$$Now as $2 - u^{2} >0$, there exists integer $N$ large enough such that $N (2-u^{2}) > 2u+1$. In which case this $N$ is such that $(u+ \frac{1}{N})^{2} < 2$, so $u+ \frac{1}{N} \in A$. But $u =\sup A < u + \frac{1}{N}$, a contradiction. So $u^{2} \not < 2$.
Similarly, suppose $u^{2} > 2$ instead. Then Consider $u - \frac{1}{N}$ for some positive integer $N$. Note again $$
\left( u - \frac{1}{N} \right)^{2} = u^{2} - \frac{2u}{N} + \frac{1}{N^{2}} > u^{2} - \frac{2u}{N}
$$Now, as $u^{2} - 2 > 0$, we can pick $N$ large enough such that $N (u^{2}-2) > 2u$. This gives $N$ such that $(u - \frac{1}{N})^{2} > 2$. But since $u = \sup A$, and $\frac{1}{N} > 0$, there exists some $r \in A$ such that $u - \frac{1}{N} < r \le u$. But this implies $r^{2} > (u- \frac{1}{N})^{2} > 2$, contradicting $r \in A$. $\blacksquare$
By the way, this subtly uses the fact that [[supplemental/Q-is-archimidean|rationals is Archimedean]], that some multiple of a positive rational will exceed other rational. ///
The real axioms assert an ordered that is Dedekind complete, which we call the reals. This Dedekind complete ordered field can be constructed, and is essentially unique up to order-isomorphism. So we will refer to it as the reals and denote it $\mathbf{R}$.
Real analysis, in one view, is to study the consequences of the reals. In particular we have all the familiar calculus theorem as a consequence of the reals. Perhaps more interestingly, one can ask the *reverse question* -- if $R$ is an ordered field such that some calculus theorem $P$ is true, what property must $R$ satisfy? And as it turns out, for many calculus theorem $P$, $R$ must be Dedekind complete, and hence the reals! So the theory of calculus is essentially the study of $\mathbf{R}$.
## Archimedean property and density of $\mathbf{Q}$.
Whenever we speak of the reals $\mathbf{R}$, we should think of it as *the* Dedekind complete ordered field. A main property of a Dedekind complete ordered field is that it has the Archimedean property, that is
> Archimedean Property.
>
> Let $a,b$ be any two numbers in $\mathbf{R}$, where $b > 0$. Then there exists some positive integer $N$ such that $Nb > a$.
Note, [[supplemental/Q-is-archimidean|the rationals is itself Archimedean]], and it is independent of the fact that $\mathbf{R}$ is as well.
To see the reals have the Archimedean property, consider the set of all positive integer multiples of $b$, denote it as $$
B = \{nb:n\in \mathbb{N}\}
$$ Note this set $B$ is not empty, as $b\in B$. Suppose that no positive integer multiple of $b$ would ever exceed $a$, then $a$ is an upper bound of $B$. So as $B$ is a subset of the reals, $\beta = \sup B$ exists in $\mathbf{R}$. So using our sup principle, there exists some element $nb \in B$ such that $\beta - b< nb$, as $b > 0$. But this shows $\beta < (n+1)b \in B$, contradicting that $\beta$ is an upper bound of $B$! Hence some positive integer multiply of $b$ exceeds $a$. $\blacksquare$
So a Dedekind complete ordered field (reals) must necessarily be Archimedean, and so is every subfield of it. For example $\mathbf{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b \in \mathbf{Q}\}$.
We say a subset $S$ of an linearly ordered set $L$ is **order dense** if every nonempty open interval $I=(a,b)$ in $L$ contains an element of $S$. As it turns out,
> An ordered field $F$ contains $\mathbf{Q}$ as an order dense subset if and only if $F$ is Archimedean.
Frist we need this "obvious lemma" regarding any Archimedean ordered field,
> If $F$ is an Archimedean ordered field, then every element $a \in F$ is between two consecutive integers.
/// Why the obvious lemma true? ///
Indeed, consider the set $A = \{n \in \mathbf{Z}: n > a\}$. By $F$ being Archimedean, there exists some positive integer $n$ such that $n > a$, so $A$ is not empty. Further $A$ is bounded below. Indeed, if $a \ge 0$, then each element in $A$ is greater than or equal to zero. If $a < 0$, then note by Archimedean, we have $m > -a$ for some positive integer $m$, so $- m < a$, hence $A$ is bounded below by the integer $-m$. Now by well-ordering property of integers, a bounded below nonempty subset of integers have a minimal element, so $M = \min(A)$ exists, and it is an integer. So $M-1 \not \in A$, whence $M-1 \le a$. Therefore $a \in [M-1,M)$. $\blacksquare$
Remark. If an ordered field is not Archimedean, elements need not be in between consecutive integers! ///
Now, suppose first that $F$ is an Archimedean ordered field, and take any nonempty open interval $I=(a,b)$ in $F$. We shall produce a rational number in the interval $I$. Note $b-a > 0$, so by Archimedean, we have some positive integer $N$ such that $N (b-a) > 1$. And as $F$ is Archimedean, we have some integer $M$ such that $$
M-1 \le Na < M
$$So, we have $$
Na < M \le Na + 1 < Nb
$$which upon by dividing by positive $N$, we get $$
a < \frac{M}{N} < b,
$$showing the rational number $\frac{M}{N} \in (a,b)$, and $\mathbf{Q}$ is order dense in $F$.
Conversely, assume $\mathbf{Q}$ is order dense in some ordered field $F$. Take $a,b \in F$, with $b > 0$. If $a \le 0$, then $1\cdot b > a$, done. So suppose $a > 0$ as well. Consider the interval $(0, \frac{b}{a})$, and rational $q = \frac{M}{N} \in (0, \frac{b}{a})$. This shows $\frac{1}{N} < \frac{b}{a}$, or $N b > a$, whence $F$ is Archimedean. $\blacksquare$
**Remark.** Order denseness in an ordered field gives rise to the same notion as topological denseness, under the induced order topology (where open intervals generates the open sets).
(Can also prove it by using Archimedean twice, first establish denominator, then numerator)
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An example of an ordered field that is *not* Archimedean is $\mathbf{Q}(x)$ with the ordering $f < g$ iff $f(x) < g(x)$ for all small positive $x$. To see this, take $1$ and $\frac{1}{x}$ from $\mathbf{Q}(x)$. Note that for any positive integer multiple of $1$, $N\cdot1 = N$, we always have $N < \frac{1}{x}$ in this ordering, as for all small enough positive $x$, we have $\frac{1}{x}$ exceeding $N$. In particular, $\frac{1}{x}$ is larger than all integers in $\mathbf{Q}(X)$! So $\mathbf{Q}(x)$ is not Archimedean, and hence cannot be Dedekind complete.
Observe that as $\frac{1}{x}$ is larger than any positive integer, it is not between any two consecutive integers, and that $\mathbf{Q}$ is not dense in $\mathbf{Q}(x)$!
In a non-Archimedean ordered field, an element larger than all positive integers $N$ is an *infinite element* or *transfinite element*, and an element smaller than all reciprocals of positive integers, $\frac{1}{N}$ is an *infinitesimal element*. So $\frac{1}{x}$ is an infinite element in $\mathbf{Q}(x)$, while $x$ would be an infinitesimal element. These infinite and infinitesimal elements do not exist in an Archimedean ordered field.
---
As one last thought on Archimedean-ness of ordered fields, as it turns out that the reals is "as Archimedean as an ordered field can be". That is, the reals is a maximally Archimedean ordered field, there is no Archimedean ordered field extending the reals:
> If $F$ is an ordered field that contains the reals as a proper subfield, then $F$ cannot be Archimedean.
///Why?///
Suppose to the contrary that $F$ is an Archimedean ordered field that contains $\mathbf{R}$ as a proper subfield. Take $x \in F - \mathbf{R}$.
Since $F$ is Archimedean, there exists some integer $N$ such that $N > x$. Now, consider the set $$
E = \{r \in \mathbf{R}: r < x\}.
$$Note $E$ cannot be empty, since if $E$ empty, this means every real number is larger than $x$, making the open interval $(x-1,x)$ disjoint from $\mathbf{R}$, and hence disjoint from $\mathbf{Q}$. But since $F$ Archimedean, $\mathbf{Q}$ is dense in $F$ and must intersects $(x-1,x)$, a contradiction. So $E \neq \varnothing$. Note also $E$ is bounded above by a real number, in particular $N$. Hence $E$ as a subset of $\mathbf{R}$, $u = \sup_{\mathbf{R}} E$ exists in $\mathbf{R}$.
Note $x \neq u$, as $x$ is not real.
If $x < u$, then by denseness of $\mathbf{Q}$, there exists some rational $q$ such that $x < q < u$. Since $u = \sup_{\mathbf{R}}E$, there exists some element $e \in E$ such that $q < e \le u$. Since $e \in E$, we have $e < x$, and hence $q < e < x$, making $q\in E$. But this contradictions $x < q$. Hence $x \not < u$.
If $x > u$, again by denseness of $\mathbf{Q}$, there exists some rational $q$ such that $u < q < x$. Note this means $q \in E$. But $u = \sup_{\mathbf{R}}E$, contradicting $q > u$. Hence $x \not > u$.
But this implies $x$ is neither equal, greater, nor less than $u$, an impossibility in an ordered field. We conclude that $F$ cannot be Archimedean! $\blacksquare$ ///
Furthermore, [[supplemental/every-archimedean-ordered-field-is-real|every Archimedean ordered field embeds into the reals]]. Hence the reals are "the most Archimedean ordered field"!
## Uniqueness and existence of the Dedekind complete ordered field.
Any two Dedekind complete ordered field are order isomorphic to each other.
A sketch of this is first note that any Dedekind complete ordered field would have an isomorphic copy of $\mathbf{Q}$ that is dense in them. Then use them to establish a field isomorphism between the two.
And such a Dedekind complete ordered field can be constructed, usually out of the rationals, such as infinite decimal expansion, Dedekind cuts, equivalence classes of Cauchy sequences. There is a way to construct the reals using just the integers, called Eudoxus reals. Interested reader can look it up.
As it might be useful, we provide Dedekind cuts as a model of the real here.
A **Dedekind cut** in an ordered field $F$ is a partition $(a,b)$ of $F = a \cup b$ such that (1) $a$ is not empty nor all of $F$, (2) $a$ is a downward ideal (if $x \in a$ and $y < x$, then $y \in a$), and (3) $a$ has no maximal element, namely for each $x \in a$, there exists $z \in a$ such that $z > x$.
Note the subset $a$ in a Dedekind cut $(a,b)$ completely determines $b$, so it is often enough to use $a$ to represent a Dedekind cut.
Now we define the reals $\mathbf{R} = \{a : a\text{ is a Dedekind cut of the rationals}\}$. Intuitively a Dedekind cut of the rationals is the set of all rationals strictly less than some real number. Each cut determines a unique real number.
Now, $\mathbf{R}$ forms an ordered set by defining $a \le b$ iff $a \subset b$. This ordering is linear. $\blacktriangleright$ Take two distinct cuts $a\neq b$, then there exists an element in one set that is not in the other. Say $x\in a -b$. Then for each $y \in b$, note $y\neq x$. If $y > x$, then as $b$ is closed downwards we have $x \in b$, a contradiction. Hence we must have $y < x$. And by $a$ is closed downwards, we have $y\in a$. This shows $b \subset a$. Symmetrically if there is an element $x' \in b -a$, then $a \subset b$. Hence this ordering is total and gives us a linearly ordered set. $\blacksquare$
A copy of the rationals in $\mathbf{R}$ can be identified as the set of all cuts of the form the open interval $(0,q)$ for each $q \in \mathbf{Q}$.
Define the field arithmetic on $\mathbf{R}$ as follows.
- Addition. $a + b :=\{x+y:x\in a,y\in b\}$.
- Subtraction. $a-b :=\{x-y : x\in a,y\in \mathbf{Q}-b \}$.
- Additive inverse. $-a:=0-a$.
- Multiplication. If $a,b \ge 0$, define $a\cdot b:=\{xy:x\in a,y\in b\}$. If $a$ or $b$ is negative, then define $a\cdot b = (-a)\cdot(-b)=-((-a)\cdot b) = -(a \cdot (-b))$ and use definition of the both positive case.
- Division. If $a\ge 0$ and $b > 0$, define $a / b := \{x / y:x \in a, y\in \mathbf{Q}-b\}$. If $a$ or $b$ negative, then define $a/b:=((-a) / (-b)) = - (-a) / b = - a / (-b)$.
- Supremum. If $S$ is a nonempty collection of cuts that is bounded above, then take $u = \bigcup_{x\in S} x$, which we claim is a cut that is the least upper bound of $S$.
One needs to verify above satisfies the field axioms, and that $\mathbf{R}$ is indeed Dedekind complete. The advantage of using Dedekind cuts is that each cut is one unique real, so we don't have to worry about equivalence classes (unlike the decimal expansion, Cauchy sequences, or the Eudoxus reals approach.)
Next installment on some consequences of the reals.