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I would like to order a group of things by a set of rules of decreasing precedence. Please critique this sentence to help illustrate that:

We define a lexicographical ordering for sheep by looking at the following properties in decreasing precedence:

  • Age in years (rounded down), lower first
  • Number of legs, higher first
  • First name in dictionary order
  • Average length of fur
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    $\begingroup$ Where is the problem/question? $\endgroup$ May 12, 2010 at 0:09
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    $\begingroup$ The question seems to be, what's the terminology you'd use for specifying such an order? I'm not sure there is any short way of saying it, though; the above seems an OK description. $\endgroup$ May 12, 2010 at 8:04
  • $\begingroup$ There is a confusion between Lexicographical and Lexical. Among many other troubles, the first name entry is not a well defined entry because of the variable length in a lexicographical order where elements are concatenated. $\endgroup$
    – ogerard
    May 13, 2010 at 7:29

1 Answer 1

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The lexical order of a well ordered list of orders $\langle X_\alpha,\leq_\alpha\rangle$ is the order on the product space $\Pi_\alpha X_\alpha$ placing $s$ before $t$ if $s_\alpha\lt_\alpha t_\alpha$ on the least difference coordinate.

But this is not your situation, since all your orders live on the same set (the set of sheep), and you seek an order on that set, not on the product set. That is, you want to order the sheep, not sequences of sheep. But if you simply restrict the lexical order to the constant sequences, you get what you want, and for this reason, I think it is fine to call it the lexical order.

Another difference is that you don't actually have orders, but pre-orders, since two sheep can have the same age, and so on. In this general setting, you have a set $X$ with a well-ordered sequence of pre-orders $\leq_\alpha$, and you define $a\leq b$ if the least index $\alpha$ for which $a$ and $b$ are not equivalent (if any such $\alpha$ exists) has $a\lt_\alpha b$.

It is important that the list of orders is well-ordered, since otherwise there may be no such least difference coordinate $\alpha$, and the idea will run aground. But any finite list of criteria is of course well-ordered, as is any $\omega$-sequence of orders.

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