What is a usual order relation?

Question

I've just started learning about relations and now I'm at partial order relations and total order relations; essentially, I'm trying to convey that I'm very much a beginner to this relations stuff.

My textbook includes the following remark:

The usual order relation on the real line $ \Bbb R $ is a total order relation.

A little later in my textbook is the following exercise (a portion of it, anyways):

Let $A$={1,2}.
List all the partial order relations on A. (The usual order relation on A is $R$={$\mathsf (1,2) \cup E)$}, where $E$={$(1,1),(2,2)$} is the relation of equality on A. ...)

The textbook also refers to a usual order relation again, somewhere later, so I would like to know what is meant by the term. I tried to infer something from the exercise whose excerpt I included, but I still don't understand. I went to mathworld.com and I didn't find anything there.

Answer 1

The "usual order relation" is a creation of laziness. It's the order relation that your intuition comes up with first when you have to define an order relation on the space it's attached to.

So for $\Bbb R$, it's the ordering of numbers, for $\{1,2\}$ it's the only "sensible" ordering "up to renaming of the elements" (i.e., $1 < 2$). I hope that clears the air.

Answer 2

The usual ordering on $\Bbb{R}$ can be uniquely defined as the total order $\lt$ on $\Bbb{R}$ that satisfies the following two properties:

(Let $a,b,c \in \Bbb{R}$)

1) $c \gt 0 \iff c$ is positive.

2) $a \lt b, \ \ 0 \lt c \implies ac \lt bc$

Proof that this defines uniquely $\lt$: Suppose that $\lt'$ also satisfies these properties and is a total order on $\Bbb{R}$. Then if $a \lt b$, assume that $b \lt' a$. Then we have $0 \lt' a-b$, or by (1) that $a-b$ is positive. Then let $c = a-b \gt' 0$. This connects us to $\lt$ by (1) since we can also then write $c = a-b \gt 0$. Now we need one more defining property to complete the proof:

3) $a \lt b \implies a + c \lt b + c$

$\square$.