Let $<_1$ be a total order on a non-empty set $A$.
Consider the set $\mathscr{J}(A)$ given by
$$\mathscr{J}(A) = \bigsqcup_{n=1}^\infty A^n$$
...where $\sqcup$ denotes "disjoint union" (or $\mathbf{Set}$'s coproduct, if you prefer), and $A^n$ is the $n$-fold Cartesian product of $A$ with itself.
For $n > 1$, let $<_n$ denote the lexicographic (total) ordering induced by $<_1$ on the subset $A^n \subset \mathscr{J}(A)$.
Since the union defining $\mathscr{J}(A)$ is disjoint, for each $x \in \mathscr{J}(A)$ there is exactly one positive integer $\Gamma(x)$ such that $x \in A^{\Gamma(x)}$.
For any two $x, y \in \mathscr{J}(A)$, define $x < y$ to mean that either
- $\Gamma(x) < \Gamma(y)$; or
- $\Gamma(x) = \Gamma(y) := \gamma$ and $x <_\gamma y$.
It's not hard to see that $<$ is a total ordering on $\mathscr{J}(A)$.
Does this construction, or any aspect of it, have a name?
Alternatively, is there a generic name for the ordering on a disjoint union of ordered sets that results when the family of disjoint subsets in the union is itself an ordered set?
(FWIW, I want to search the literature on algorithms for working with the sequence of elements in $\mathscr{J}(A)$ induced by the total ordering $<$, for a given finite $A$.)
You're describing the indexed sum of linear orders: if $\mathcal{L}_i=(A_i, \prec_i)$ is a family of linear orders for $i\in I$, and $\prec_I$ is a linear order on $I$, then the indexed sum (or coproduct) of the $\mathcal{L}_i$s along $I$ is the set $$\{(i, a)\in I\times\bigsqcup_{i\in I} A_i: a\in A_i\},$$ ordered by $$(i, a)<(j, b)\iff [i<_Ij\mbox{ or } (i=j\mbox{ and } a<_ib)].$$
In your specific example, ($I, \prec_I)=(\mathbb{N}, <)$ and $\mathcal{L}_i=(A^i, <_i)$. See e.g. https://en.wikipedia.org/wiki/Total_order#Sums_of_orders.
In describing this, by the way, I would use language rather than symbols (or at least, in addition to symbols); e.g. "Ordered first by length, and second lexicographically." It gets the point across and the formal description of even a reasonably simple indexed sum can be quite messy.