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A lot has been written about the arithmetic of ordinal numbers. However, we can also do arithmetic with linearly ordered sets.

Question. Is there an article or book where I can learn the basics of linearly ordered set arithmetic?

Here's the definitions I have in mind. Let $I$ denote a totally ordered index set, and suppose $L$ is an $I$-indexed family of linearly ordered sets. Then:

Definition 0. The summation $\sum_{i \in I} A_i$ is defined in the same way as for ordinals; in particular, it is the linearly ordered set consisting of all the $A_i$'s stuck head-to-tail in the order they're indexed. Further, we define $A \times B$ as shorthand for $\sum_{b \in B} A$ whenever $A$ and $B$ are totally ordered sets.

Definition 1. The product $\prod_{i \in I}A_i$ is defined only if $I$ is in fact well-ordered, in which case it is the linearly ordered set whose underlying set is the obvious product of sets, and whose order relation is defined lexicographically. Further, we define $A^\beta = \prod_{\alpha \in \beta} A$ whenever $A$ is a linearly-ordered set and $\beta$ is well-ordered.

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  • $\begingroup$ Note that in the slightly-narrower setting of $\{$linear orders with fixed base point$\}$, there's one more natural operation you can consider: (infinite) direct sum, with all but finitely many entries equal to the distinguished element. Unlike direct products, we can meaningfully take direct sums over merely linearly ordered sets, since "check the least index with disagreement" still works. $\endgroup$
    – Noah Schweber
    Commented Apr 2, 2014 at 6:21
  • $\begingroup$ @NoahS ah I have finally understood what you mean. True! $\endgroup$
    – goblin GONE
    Commented Apr 3, 2014 at 6:48
  • $\begingroup$ @NoahS, here's another one. If $B$ is a linear order with fixed basepoint $b$ and $L$ is an arbitrary linear order, write $B^L$ for the set of all functions $f : L \rightarrow B$ such that $\{x \in L \mid f(x) \neq b\}$ is a well-ordered subset of $L$. $\endgroup$
    – goblin GONE
    Commented Apr 3, 2014 at 6:55
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    $\begingroup$ These two can be combined: take $\bigotimes^{wf}_{L}(B_i, b_i)$ to be the set of elements $f$ of $\Pi_LB_i$ with $\{l\in L: f(l)\not=b_l\}$ a well-ordered subset of $L$. Given any two such $f_0, f_1$, there must be a first element on which they disagree, so the first-disagreement order makes sense. $\endgroup$
    – Noah Schweber
    Commented Apr 3, 2014 at 7:17
  • $\begingroup$ @NoahS, yep, cool. Furthermore, $\bigotimes^{wf}_{i \in L}(B_i,b_i)$ agrees with $\prod_{i \in L}B_i$ whenever $L$ is well-founded. $\endgroup$
    – goblin GONE
    Commented Apr 3, 2014 at 7:29

1 Answer 1

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Try Rosenstein: Linear orderings

http://books.google.com/books/about/Linear_orderings.html?id=y3YpdW-sbFsC

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