A Partial Take on Partial Order

What makes a number register in our minds? In most instances, it would be the notion of order. Here, we discuss the ordering of $\mathbb N$ , then explore it in the context of other sets like $\mathbb Z, \mathbb Q,\mathbb R$ . Remarkably, we cannot extend this notion to the complex numbers $\mathbb C$ .

What do we mean when we say that $1 \leq 2$ ? Essentially, it means that $2 = 1 + 1 = 1 + k$ for some nonnegative integer $k$ . We can formalise as follows.

Definition 1. Define the not-greater relation $\leq$ on $\mathbb N$ by

$\displaystyle x \leq y \quad \iff \quad (\exists k \in \mathbb N : y = x + k),$

where we denote $x \leq y \iff (x,y) \in\ \leq$ for readability.

Definition 2. A relation $\preceq$ on a set $K$ is a partial order if it is reflexive, transitive, and anti-symmetric:

$\forall x,y \in K\quad (x \preceq y \wedge y \preceq x \Rightarrow x = y).$

For any partial order $\preceq$ on $K$ , we write $x \prec y$ to mean $(x \preceq y) \wedge (x \neq y)$ and $x \succ y \iff y \prec x$ . A partial order is a total order if for any $x, y \in K$ , $(x \preceq y) \lor (y \preceq x)$ .

Theorem 1. The relation $\leq$ on $\mathbb N$ is a total order.

Proof. Exercise.

How do we order $\mathbb Z$ ? We use the ordering on $\mathbb N$ creatively. Recall that integers can be conceived as sets of the form $[(x,y)]$ , which is interpreted to mean, loosely, the quantity $x - y$ , whether or not this expression is meaningful:

$[(x_1,y_1)] \leq [(x_2,y_2)]\quad \iff \quad x_1 + y_2 \leq x_2 + y_1.$

Notice that on the right-hand side, we have the ordering as defined in Definition 1. In fact, particularising to $y_1 = y_2 = 0$ gives us the original formulation. We will leave its proof as a bona fide total order as an exercise in bookkeeping.

Theorem 2. Define the ordering $\leq$ on $\mathbb Z = \mathbb N^2/{\equiv}$ by

$[(x_1,y_1)] \leq [(x_2,y_2)]\quad \iff \quad x_1 + y_2 \leq x_2 + y_1.$

Then $\leq$ is a well-defined total order on $\mathbb Z$ that agrees with the ordering on $\mathbb N$ in Definition 1.

Just like with integers, we can define ordering in $\mathbb Q$ in a similar manner.

Theorem 3. Define the ordering $\leq$ on $\mathbb Q = (\mathbb Z \times \mathbb N)/{\equiv}$ by

$[(x_1,y_1)] \leq [(x_2,y_2)]\quad \iff \quad x_1 y_2 \leq x_2 y_1.$

Then $\leq$ is a well-defined total order on $\mathbb Q$ that agrees with the ordering on $\mathbb Z$ in Theorem 2.

With some effort, the ordering on $\mathbb R$ can be shown to exist and agree with the ordering in $\mathbb Q$ , though as with any notion connected with $\mathbb R$ , should be relegated to a study in real analysis.

Lemma 1. The ordering on $\mathbb Q$ satisfies the following properties:

For any $x,y,z \in \mathbb Q$ , $x \leq y \Rightarrow x+z \leq y+z$ .
For any $x,y \in \mathbb Q$ , $(x \geq 0) \wedge (y \geq 0) \Rightarrow xy \geq 0$ .

With some effort, the properties on $\mathbb Q$ also transfer to $\mathbb R$ . We call them ordered fields. More precisely, we will show that if $\mathbb C$ is an ordered field, its ordering violates the ordering on $\mathbb R$ , and thus all the orderings defined on its subsets $\mathbb Q, \mathbb Z, \mathbb N$ . For now, we can think of $\mathbb C$ as a field that contains $\mathbb R$ as a subset, together with the peculiar element $i := \sqrt{-1}$ with the property that $i^2 = -1$ .

Theorem 4. Suppose $\leq$ is a total order on $\mathbb C$ such that $\mathbb C$ is an ordered field. Then $\leq$ does not agree with $\mathbb R$ .

Proof. Suppose $\leq$ agrees with $\mathbb R$ . We will derive a contradiction. Since $\leq$ is a total order, we either have $1 \leq i$ or $i \leq 1$ . Suppose the former without loss of generality. Then