General definition of antilexicographic order on arbitrary indexed family of ordered sets Just & Weese in their "Discovering Modern Set Theory, I The Basics" (1996) defines the lexicographic order on an aribtrary indexed family of ordered sets as follows:
Let $(I, \preceq)$ be an arbitrary well-ordered set.
For each $i \in I$, let $(A_i, \preceq_i)$ be an ordered set (possibly partially).
Then $\displaystyle {\bigotimes_{i \mathop \in I} }^l (A_i, \preceq_i)$ is the ordered set $(D, \preceq_D)$ where $\displaystyle D = \prod_{i \mathop \in I} A_i$ and:
$$u \preceq_D v \iff \begin{cases} u = v \\ u(i) \prec_i v(i) \text { for the smallest $i \in I$ such that $u(i) \ne v(i)$}\end{cases}$$
I have paraphrased and reformatted for clarity.
All well and good (apart from a typo in the last line of the presentation in J&W where they say $u(i) <_i v(i)$ instead of $u(i) \prec_i v(i)$
The reason for the well-ordered nature of the indexing set is clear: you've got to have a "smallest" $i \in I$ for the definition of the ordering to be definable.
Various results are then set as exercises: prove that the lex product so defined is indeed an ordering; prove that if the ordered sets are totally ordered then so is the lex product; if the ordered sets are well-ordered then it is not necessarily the case that so is the lex product. (They do not however go on to discuss the case where $I$ is finite, as other texts sometimes do, where it is shown that the lex product of well-ordered sets is well-ordered, but never mind.)
They then set as an exercise the task of setting up the simple direct product (straightforward: the indexing set does not need even to be ordered) and the antilex product. As the question goes: "Pay special attention to the kind of order relation you need on the index set (if, in fact, you need any such relation)."
So, I gird up my loins and take the following approach:
Let $(I, \preceq)$ be an arbitrary totally ordered set. (I'm not completely sure about this -- but I have a suspicion about this which I will come to later.)
For each $i \in I$, let $(A_i, \preceq_i)$ be an ordered set.
Then $\displaystyle {\bigotimes_{i \mathop \in I} }^a (A_i, \preceq_i)$ is the ordered set $(D, \preceq_D)$ where $\displaystyle D = \prod_{i \mathop \in I} A_i$ and:
$$u \preceq_D v \iff \begin{cases} u = v \\ \exists i \in I: \forall j \succ i: u(j) = v(j) \text { and } u(i) \preccurlyeq_i v(i) \end{cases}$$
That is, if ''past a certain point'' $i$, the sequences $u$ and $v$ are equal, then the ordering is done on the element ''at that point''.
It is then immaterial whether or not there is a "smallest element" of $I$, and so $I$ need not be well-ordered. (Hence my suspicion above that it may be sufficient for $I$ to be totally ordered. I can't get my head round the concept of exactly what would happen if $I$ is only partially ordered, it needs more thought.)
But now we run into another problem. What if there is no such $i$ such that $\forall j > i: u(j) = v(j)$? What if $u$ and $v$ are such that, for example, $I = \mathbb N$ where $u(2n) \prec v(2n)$ but $u(2n+1) \succ v(2n+1)$? It seems you cannot define an antilex ordering on such a tuple.
This has repercussions. My (as yet limited) understanding is that the "order product" is defined on ordinals by means of exactly this antilex ordering. Great pains are taken in the texts to establish the properties of transfinite ordinal arithmetic based on exactly this construct, but as far as I've been able to read ahead and understand what I'm skimming through, the "order product" is rigorously explored only in the case of finite instances of such products.
I have not been able to find any discussion of the above on the net, or in my (admittedly limited) reading of the subject in works I've got (unless I study every single line carefully, and make an effort to understand all the exercises, I get lost in exposition very easily).
So: how do you define an antilex order on an arbitrary (possibly infinite) indexing set? Do you need to define an analogue of the "well-ordered set" such that its dual is well-ordered? J&W mention dual orderings briefly, but scamper on without dwelling long; is there a well-understood concept of a "dual well-ordering" which is one such that every subset has a ''greatest'' element? If so, it makes me wonder whether this is what is needed. If it really is that simple, why is this not explained and explored in the literature -- or if it is, where?
 A: 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 $i\in I$.
