4.1 Total orders and ordered fields

Comparing elements and comparing numbers

The constructions of $N$, $Z$ and $Q$ provide objects and arithmetic. Bounds ask a different question: which elements lie above or below every member of a set? Answering it requires an order relation. Two features must be distinguished: whether every pair is comparable, and whether comparison respects arithmetic. Divisibility supplies an example where the first feature fails; an ordered field will require both.

Partial order versus total order

Definition 4.1.1

(Total order)

A set $X$ with relation $\le$ is totally ordered if the following hold for all $x,y,z\in X$:

1. $x\le x$ for every $x\in X$,
2. $x\le y$ and $y\le x$ imply $x=y$,
3. $x\le y$ and $y\le z$ imply $x\le z$,
4. for every $x,y\in X$, either $x\le y$ or $y\le x$.

The first three conditions say that $\le$ is a partial order. The fourth condition is the extra comparability condition that makes the order total.

The crucial new feature is comparability. In a total order, there are no two elements left floating without comparison.

Worked example 4.1.2

(A standard total order)

The usual order on $\{1,2,3,4\}$ is total. For any two elements, one lies at or to the left of the other, so comparison is never ambiguous.

Worked example 4.1.3

(A partial order that is not total)

Let $X=\{1,2,3,6\}$ and declare $x\le y$ when $x$ divides $y$.

This is a partial order, but not a total order. The elements $2$ and $3$ are incomparable: $2$ does not divide $3$, and $3$ does not divide $2$.

This example is important because it shows that “ordered” does not automatically mean “every pair is comparable.” That extra property has to be checked.

Restricted order on a subset

Once an ambient set is totally ordered, every subset inherits that order.

Theorem 4.1.4

(Restricted order stays total)

If $(X,\le)$ is totally ordered and $Y\subseteq X$, then $Y$ becomes totally ordered when we use the same comparison rule on elements of $Y$.

For example, the set

$$S=\{q\in Q\mid q^2\lt 2\}$$

inherits its comparisons from $Q$; no new order on $S$ is needed.

The standard order on $Z$ and $Q$

For integers and rationals, the familiar order can be written in a way that fits the algebra already built in earlier chapters:

$$x\le y \quad \text{iff} \quad 0\le y-x.$$

So comparison is translated into a statement about the sign of a difference. This is useful because sign, addition, and multiplication are already part of the algebraic language of $Z$ and $Q$.

Theorem 4.1.5

(The standard order on $Z$ and $Q$ is total)

With the usual notion of positivity, the standard order on $Z$ and on $Q$ is a total order.

The reason is structural: for any difference $y-x$, exactly one of three things happens. It is positive, it is zero, or it is negative. Those three cases give $x\lt y$, $x=y$, or $x\gt y$, so every pair is comparable.

Order and operations must cooperate

The order is not an isolated decoration. It has to behave properly with the field operations.

Definition 4.1.6

(Order-compatible algebra facts)

For rational numbers $x,y,z$:

• if $x\le y$, then $x+z\le y+z$;
• if $x\ge 0$ and $y\ge 0$, then $xy\ge 0$.

The first rule says translation preserves order. The second says multiplying nonnegative quantities cannot suddenly create a negative result.

Worked example 4.1.7

(Why translation invariance matters)

If $1/3\le 1/2$, then adding $2$ to both sides gives

$$2+\frac13 \le 2+\frac12.$$

This is not a new theorem each time. It is the same structural rule applied to different numbers.

Without these compatibility rules, the order would not interact correctly with the algebra. Later arguments about intervals, bounds, and absolute values would fall apart.

Fields and ordered fields

One can summarize almost everything we know about $Q$ by packaging the operations together.

Definition 4.1.8

(Field)

A field is a set $F$ with distinct distinguished elements $0,1\in F$, addition and multiplication maps $F\times F\to F$, and the following laws for all $x,y,z\in F$:

• addition and multiplication are associative and commutative,
• $0\ne1$, and $0+x=x$, $1x=x$ for all $x\in F$,
• multiplication distributes over addition,
• every $x$ has an additive inverse $-x$ with $x+(-x)=0$,
• every $x\ne0$ has a multiplicative inverse $x^{-1}$ with $xx^{-1}=1$.

Closure is part of the operation-map statement: adding or multiplying two elements of $F$ must return an element of $F$.

This definition remembers the algebra, but it says nothing about order.

Definition 4.1.9

(Ordered field)

An ordered field is a field $F$ equipped with a total order $\le$ such that, for all $x,y,z\in F$:

• $x\le y$ implies $x+z\le y+z$,
• $x\ge 0$ and $y\ge 0$ imply $xy\ge 0$.

So an ordered field is not just “a field plus a comparison symbol.” It is a field whose algebra and order fit together coherently.

The rational numbers $Q$ are an ordered field. The real numbers $R$ will also be an ordered field, but chapter 4 will show that $R$ has one more crucial property beyond this: completeness.

Common mistake

(A total order is more than a left-to-right picture)

Students often think a total order is just “something that can be drawn on a line.” The real point is the comparability axiom. A partial order can still have a clear diagram or hierarchy and yet fail to compare some pairs.

Common mistake

(A field is not automatically an ordered field)

The field axioms only describe algebraic operations. To become an ordered field, the set also needs a total order, and that order must respect addition and multiplication in the precise way stated above.

Comparing orders by the hypotheses they support

There are two questions that are often compressed into the word “order.” The first asks whether two elements can be compared. The second asks whether that comparison is compatible with arithmetic. The first is answered by a relation; the second requires extra structure.

Definition 4.1.10

(Strict order and non-strict order)

In a partially ordered set, write $x\lt y$ when $x\le y$ and $x\ne y$. In a total order, exactly one of $x\lt y$, $x=y$, or $y\lt x$ holds for every pair. In an ordered field, the strict relation is also compatible with translation and positive scaling: $x\lt y$ implies $x+z\lt y+z$, and $x\lt y$ together with $0\lt z$ implies $xz\lt yz$.

The distinction explains why the same-looking argument can be valid in $Q$ but invalid in a set ordered by divisibility. In $Q$, the expression $(y-x)/2$ is available and positivity can be transported through it. In a divisibility order, “halfway between” two elements is not an order-theoretic notion at all.

Worked example 4.1.11

(A translation argument in an ordered field)

Assume $a\lt b$ and $c\lt d$ in an ordered field. Adding $c$ to the first inequality gives $a+c\lt b+c$; adding $b$ to the second gives $b+c\lt b+d$. Transitivity yields $a+c\lt b+d$. Notice the proof uses only translation invariance and transitivity; it does not rely on a picture of the number line.

Worked example 4.1.12

(A positive-product argument)

If $0\lt a\lt b$ and $0\lt c\lt d$, then $ac\lt bc$ because multiplying by the positive $c$ preserves strict order, and $bc\lt bd$ because multiplying by the positive $b$ preserves the second inequality. Thus $ac\lt bd$. The sign hypotheses are not cosmetic: multiplying by a negative number reverses an inequality.

The order laws in proof form

The definitions become useful only when we can turn them into reliable proof steps. For $x,y\in Q$, exactly one of $y-x\gt0$, $y-x=0$, or $y-x\lt0$ holds. Using $x\le y$ precisely when $0\le y-x$, these three cases give $x\lt y$, $x=y$, or $y\lt x$. This proves trichotomy and hence comparability for the standard order. It also shows why the order is antisymmetric: if both $x\le y$ and $y\le x$, then both $y-x$ and $x-y$ are nonnegative, so their sum is zero and therefore $x=y$.

Here is the translation proof in its general form. If $x\le y$, then $0\le y-x$. For any $z$,

$$(y+z)-(x+z)=y-x,$$

so the same nonnegative difference gives $x+z\le y+z$. Conversely, applying this rule with $-z$ shows that $x+z\le y+z$ implies $x\le y$; translation is therefore an order isomorphism. In particular, subtraction is not a separate order rule: it is translation by an additive inverse.

Positive scaling requires a sign hypothesis. If $x\le y$ and $0\le z$, then $0\le y-x$ and multiplication of nonnegative rationals gives $0\le z(y-x)=zy-zx$; hence $zx\le zy$. If $z>0$ and $x<y$, then $y-x>0$, so $z(y-x)>0$ and the inequality is strict. If $z<0$, write $z=-w$ with $w>0$; positive scaling by $w$ followed by negation reverses the inequality. This is the algebraic reason that cancellation by a positive number preserves order, whereas cancellation by a negative number reverses it.

Theorem 4.1.13

(Trichotomy and order compatibility in Q)

For every $x,y\in Q$, exactly one of $x\lt y$, $x=y$, and $y\lt x$ holds. Moreover, translation by any rational preserves both $\le$ and $<$, while multiplication by a positive rational preserves them and multiplication by a negative rational reverses them.

These statements also clarify the quantifiers used later. The assertion $x\le y$ is a statement about one pair. The assertion that $u$ is an upper bound of $Y$ is stronger:

$$(\forall y\in Y)\;y\le u.$$

To refute it, one must produce a single witness $y\in Y$ with $u\lt y$ in a total order. A picture with several points below $u$ cannot establish the universal claim, and a list of examples cannot refute it without such a witness.

Worked example 4.1.14

(Three subsets, three different order questions)

In $X=\{-2,0,3\}$ with its usual total order, $Y=\{-2,3\}$ has minimum $-2$ and maximum $3$; its upper bounds in $X$ are just $3$, while its upper bounds in $Q$ are every rational $u\ge3$. The interval $Z=(0,1)$ has no maximum, although $1$ is an upper bound in $Q$; for every $u\lt1$, a point of $Z$ lies above $u$. Finally, $Q$ is unbounded above, because $u+1$ defeats every candidate $u$. The ambient ordered set and the quantifiers change the answer.

For example, an interval such as $[0,1]\subseteq R$ inherits a total order, but it is not a field: $2$ is not in the interval, and a positive element such as $1/2$ has no additive inverse in the interval. We may discuss bounds and suprema of subsets of the interval, but we must name $R$ as the ambient ordered field when invoking completeness or subtraction.

Common mistake

(Closure is separate from inherited order)

Every subset inherits comparisons, but it need not inherit algebraic closure. Do not call an interval a field merely because it sits inside $R$; state the ambient field whenever an argument uses its operations.

A compact proof toolkit for ordered fields

Several consequences of the ordered-field axioms recur so often that they are worth proving once. The multiplicative axiom initially gives only $a\ge0$ and $b\ge0$ implying $ab\ge0$; strict positivity must be derived. First prove $0\lt1$. The field axiom gives $1\ne0$. If $1\lt0$, translation by $-1$ gives $0\lt-1$, so $-1\ge0$. The nonnegative-product axiom then gives

$$0\le(-1)^2=1,$$

contradicting $1\lt0$. Trichotomy therefore gives $0\lt1$. If $a\lt b$, adding $-a$ gives $0\lt b-a$, so every strict inequality can be rewritten as positivity of a difference.

Now suppose $0\lt a$ and $0\lt b$. The axiom gives $ab\ge0$. A field has no zero divisors: if $ab=0$, then multiplying by $a^{-1}$ or $b^{-1}$ would force one of $a,b$ to be zero. Hence $ab\ne0$, and trichotomy upgrades $ab\ge0$ to $0\lt ab$. In particular, if $a\ne0$, then $a^2\ge0$ and $a^2\ne0$, so $a^2\gt0$. This gives the useful implication

$$a^2=0\quad\Longrightarrow\quad a=0.$$

The inverse of a positive element is positive as well. If $c\gt0$, then $c^{-1}\ne0$ because $cc^{-1}=1$. It cannot be negative: if $c^{-1}\lt0$, then $-c^{-1}\gt0$, and the strict-product result would give $c(-c^{-1})=-1\gt0$, contradicting $-1\lt0$. Thus $c^{-1}\gt0$. Now cancellation is justified: if $c\gt0$ and $ac\le bc$, then $0\le(b-a)c$; multiplying by the positive inverse $c^{-1}$ preserves nonnegativity and gives $0\le b-a$, hence $a\le b$. This is why inequalities involving fractions must track denominator signs. For nonzero $b,d$, cross-multiplying $a/b<c/d$ preserves direction when $bd>0$ and reverses it when $bd<0$. Taking both denominators positive is one way to ensure $bd>0$.

Worked example 4.1.15

(Checking an inequality by its difference)

To prove $-2/3\lt5/6$, do not rely on a diagram. Compute

$$\frac56-\left(-\frac23\right)=\frac56+\frac46=\frac96=\frac32\gt0.$$

The positivity of this difference is exactly the definition of the strict order. If both sides are translated by $7$, the difference remains $3/2$, so $19/3\lt47/6$ follows from the same proof.

These elementary implications are the bridge from the abstract axioms to later bound arguments. They justify taking midpoints such as $(a+b)/2$ in an ordered field, but they do not justify taking a midpoint in an arbitrary partial order. The distinction will matter when a supremum argument is transferred from a total order to a partial order.

Quantifiers determine the order claim

Order notation often hides the domain over which a statement is quantified. The claim $x\lt y$ is local, but an interval statement is universal. For example,

$$(\forall x\in(0,1))\;x<1$$

proves that $1$ is an upper bound of $(0,1)$, while

$$(\forall u<1)(\exists x\in(0,1))\;u<x$$

proves that no smaller real number is an upper bound. The order alone does not turn one of these formulas into the other; the second needs an explicit witness, such as $x=(u+1)/2$ when $u\ge0$, or $x=1/2$ when $u<0$.

The same discipline handles lower bounds. To prove that $0$ is the infimum of $(0,1)$, show $0\le x$ for every $x$ in the interval, then take any $v>0$ and choose a point of the interval below $v$. The choice $x=\min\{v/2,1/2\}$ works because it is positive, below $1$, and strictly below $v$. These witness formulas are the elementary form of the approximation criteria developed in the next note.

Empty and unbounded domains expose the limits of the definitions. For $Y=\varnothing$, every $u$ is both an upper and a lower bound, since a universal statement over no elements is true. Thus a claim about a “best” bound for the empty set needs a separate convention and is excluded from the usual completeness axiom. For $Y=Q$, every proposed upper bound $u$ is defeated by $u+1$, and every proposed lower bound by $u-1$; the set is unbounded in both directions. The quantifiers are what make these conclusions rigorous.

Worked example 4.1.16

(One interval, four quantified statements)

For $Y=[-1,2]$ in $R$, $2$ is an upper bound because every $y\in Y$ satisfies $y\le2$, and it is a maximum because $2\in Y$. The element $3$ is also an upper bound but is not a maximum because $3\notin Y$. The number $-1$ is a minimum and an infimum. If the interval is changed to $(-1,2)$, the same bound inequalities remain true, but neither endpoint is a member; therefore there is no maximum or minimum. Changing one strict symbol changes membership without changing the surrounding order structure.

These examples illustrate a general proof habit. Begin by naming the ambient ordered set, because the same subset can have different available bounds after the universe changes. Next write the statement with its quantifiers. A maximum claim contains an existential membership clause and a universal comparison clause; an upper-bound claim contains only the universal comparison clause. Finally, when proving that a bound is sharp, state how a proposed improvement is defeated. For an upper bound this means finding a set element above the proposal; for a lower bound it means finding a set element below it.

The algebraic axioms support these witnesses in an ordered field. Midpoints are available because division by $2$ is defined and $2>0$. Positive scaling allows an inequality to be transported through multiplication, and additive inverses allow every translation argument to be reversed. None of these facts creates a supremum. They only let us verify a candidate once one has been identified. The existence of a least upper bound for every nonempty set bounded above is the separate completeness property studied after the definitions of bounds.

Quick checks

Quick check

(Why is divisibility on positive integers not a total order?)

Look for two incomparable elements.

Solution · Answer

Because some pairs cannot be compared. For example, $2$ does not divide $3$, and $3$ does not divide $2$, so the comparability condition fails.

Quick check

(If $x\le y$, why does $x-z\le y-z$ also hold?)

Rewrite subtraction as addition.

Solution · Answer

Because $x-z = x+(-z)$ and $y-z = y+(-z)$. The order-compatibility rule for addition says adding the same number to both sides preserves the inequality.

Exercises

Quick check

(Explain why every subset of a totally ordered set inherits a total order.)

Take two elements of the subset and compare them in the ambient set.

Solution · Guided solution

Let $y_1,y_2\in Y\subseteq X$. Since $X$ is totally ordered, either $y_1\le y_2$ or $y_2\le y_1$ in $X$. The restricted order on $Y$ uses the same comparison, so the same alternative holds in $Y$. The other order axioms were already true in $X$, so they remain true on the subset.

Quick check

(Why is $Q_{\gt 0}$ not a field?)

Check closure and inverses against the field axioms.

Solution · Guided solution

The positive rationals are closed under multiplication and have multiplicative inverses, but they do not have additive inverses inside the same set. If $q\gt 0$, then $-q$ is not in $Q_{\gt 0}$. So the field axioms fail.

Related notes

Read this after 3.4 Rationals and well-defined operations and 3.5 Gaps in Q and sqrt(2). Then continue to 4.2 Upper bounds, supremum, and infimum.