Here is a simple example to demonstrate that a field together with an ordering on the underlying set is not automatically an ordered field.
The field is $\Bbb R$ with usual addition and multiplication.
The ordering is as follows. Let $\ll$ denote the following relation: $x \ll y$ if $x$ and $y$ are both rational and $x<y$ by the usual order, or if $x$ and $y$ are both irrational and $x<y$ by the usual order, or if $x$ is rational and $y$ is irrational.
In this order, we have both $\pi \gg 0$ and $-\pi \gg 0$, but their sum is not $\gg 0$.
Let me deal with ordinal exponentiation first. For ordinals $\alpha$ and $\beta$ and a function $f:\beta\to\alpha$ define the support of $f$ to be $\operatorname{supp}(f)=\{\xi\in\beta:f(\xi)\ne 0\}$. Let
$$F=\left\{f\in{}^\beta\alpha:\operatorname{supp}(f)\text{ is finite}\right\}\,;$$
if $f,g\in F$ and $f\ne g$, there is a largest $\delta_{f,g}\in\beta$ such that $f(\delta_{f,g})\ne g(\delta_{f,g})$. For arbitrary $f,g\in F$ we set $f\preceq g$ iff either
- $f=g$, or
- $f\ne g$ and $f(\delta_{f,g})<g(\delta_{f,g})$.
This is essentially your definition of the anti-lexicographic order on ${}^\beta\alpha$, but limited to the subset of functions with finite support, so that there always is a last index at which distinct elements of $F$ differ. It is this $\langle F,\preceq\rangle$ that is order-isomorphic to the ordinal power $\alpha^\beta$.
Ordinal multiplication is more straightforward, since we’re dealing with a product of only two orders: the ordinal product $\alpha\cdot\beta$ is the order type of the anti-lexicographic order on $\alpha\times\beta$, i.e., of the lexicographic order on $\beta\times\alpha$. Informally, it’s the concatenation of $\beta$ copies of $\alpha$.
I don’t know exactly what Just and Weese have in mind, but one way to define the anti-lexicographic order on a product of orders is as follows.
Let $\langle I,\preceq\rangle$ be an arbitrary anti-well-ordering, so that $\preceq$ is a linear order on $I$, and every non-empty subset of $I$ has a $\preceq$-greatest element. For each $i\in I$ let $\langle A_i,\preceq_i,a_i\rangle$ be a partial order with a base point $a_i$. Let $D=\prod_{i\in I}A_i$, and for $f\in D$ let
$$\operatorname{supp}(f)=\{i\in I:f(i)\ne a_i\}\,.$$
Clearly either $f(i)=a_i$ for each $i\in I$, or $\operatorname{supp}(f)$ has a $\preceq$-largest element. Thus, if $f,g\in D$, and $f\ne g$, there is a $\preceq$-largest $i_{f,g}\in I$ such that $f(i_{f,g})\ne g(i_{f,g})$. For arbitrary $f,g\in D$ we set $f\preceq_Dg$ iff either
- $f=g$, or
- $f\ne g$ and $f(i_{f,g})\preceq_{f,g}g(i_{f,g})$.
This is especially nice if $a_i$ is the $\preceq_i$-minimum element of $A_i$ for each $i\in I$.
Best Answer
You're touching on the field of "order-theory" (which has connections to model theory, topology and set theory, mostly). The axioms describe a (strict) linear order, and the standard examples are $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ which almost anyone doing mathematics encounters somehow. All these have a standard linear order associated with them.
In general we consider two linearly ordered sets $(X,<_X)$ and $(Y,<_Y)$ to be order-isomorphic when there exists a bijection $f: X \to Y$ such that $$\forall x,x' \in X: x <_X x' \iff f(x) <_Y f(x')$$
and such sets are considered to have "the same order structure". Abstractly we have a partition into linearly ordered sets into "equivalence class" of this with the same order, and such an equivalence class is called an "order type". We can talk about the order type of $\mathbb{N}$, as all linear orders that are order isomorphic to $\mathbb{N}$ (intuitively: start with a minimal element, and add a unique successor for every element, and don't stop (or unit we first have an infinite number of elements). A boring example: the even numbers in their inherited order are the same order type as $\mathbb{N}$, and some thought will show that the same holds for all infinite subsets of $\mathbb{N}$. For every finite number $n$ there is a unique order type of linear orders of type $n$, but this is definitely not the case for countable sets.
Some order properties for some countable orders that can serve to show they're not order-isomorphic:
$\mathbb{N}$ has a minimum, $\mathbb{Z}$, $\mathbb{Q}$ do not.
$\mathbb{Z}$ has the property that every point has two neighbours (a right neighbour of $x$ is an $x^+$ such that $x < x^+$ but not $x'$ exists with $x < x' < x^+$; a left neighbour is the same but smaller), while $\mathbb{Q}$ does not have that property and in $\mathbb{N}$, $0$ has no left neighbour.
$\mathbb{Q}$ is order dense ($\forall x,y \in X: (x < y) \to (\exists z \in X: x < z < y)$ but the orders $\mathbb{N}$ and $\mathbb{Z}$ are not (see previous property).
It turns out that the order type of $\mathbb{Q}$ is special: If $(X,<)$ is a countable linearly ordered set, and $X$ is order dense and has no maximum or minimum, then $X$ is order isomorphic to $\mathbb{Q}$, and moreover any countable linearly ordered set is order-ismorphic to some subset of $\mathbb{Q}$ (in the inherited order from the rationals). So all countable can certainly be embedded into a number line (namely $\Bbb Q$), but for larger cardinals this need not be true, I believe. For a little more on order types see its Wikipedia page, e.g.
The well-ordered (meaning: every non-empty subset has a minimum) linear orders get a special study in set theory. They are uniquely represented by so-called ordinal numbers. They form the basis for cardinal numbers.
And $\Bbb Q$ can be so-called "order-completed" (there are still gaps in it: sets with upper bound but with no least upper bound) and it turns out the order type of the order-completion of $\Bbb Q$ is exactly $\Bbb R$ and in analysis this order compeleteness is used all over the place.
A standard work on linear order theory see the book "Linear Orderings", by Rosenstein. You can define addition and multiplication of order types e.g. Any linearly ordered set has a natural topology too (in fact for all four examples in the first paragraph, their natural topologies are in fact the order topology), that I recently talked about here, if you're interested.