Thematics Of Order

[Jon Awbrey]

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TOO.  Note 6

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Re: TOO 5.  http://stderr.org/pipermail/inquiry/2005-April/002551.html
In: TOO.    http://stderr.org/pipermail/inquiry/2005-April/thread.html#2541

We are looking for maps of the form M^k -> M, where M = {1, 2, 3, ...},
that give us some measure of the orderliness in a k-tuple <x_j> of M^k,
in other words, a measure of how true is the statement about the order,
or the chain of inequalities, averring x_1 < x_2 < ... < x_(k-1) < x_k.

Let us first consider k-tuples all of whose elements are distinct.
This makes sense inasmuch as a k-tuple with repeating elements is
essentially a tuple of a lower order, and a sensible tactic would
be to treat it as already covered.  For example, we could include
our measure in a function of the form M^k -> B x M, assigning any
k-tuple with repeating elements the character 0 in B.  But we can
leave this decision for another time.

Here is one idea of how to find a measure of order m : M^k -> M.
Let p_j be the j^th prime.  Thus p_1 = 2, p_2 = 3, p_3 = 5, etc.
Given a k-tuple <x_1, ..., x_k> in M^k, in other words, a finite
sequence x of length k over M, or a function x from the interval
[1, k] to M, written x : [1, k] -> M, we define m(x) as follows:

   m(x)  =  (p_1)^x(1) * ... * (p_k)^x(k)

         =  (p_1)^x_1 * ... * (p_k)^x_k

By way of a concrete example, let us consider the six 3-tuples in M^3
that can be formed by permuting the elements of the 3-tuple <1, 2, 3>.

   m(<1, 2, 3>)  =  2^1 * 3^2 * 5^3  = 2 *  9 * 125  =  2250

   m(<2, 1, 3>)  =  2^2 * 3^1 * 5^3  = 4 *  3 * 125  =  1500

   m(<1, 3, 2>)  =  2^1 * 3^3 * 5^2  = 2 * 27 *  25  =  1350

   m(<3, 1, 2>)  =  2^3 * 3^1 * 5^2  = 8 *  3 *  25  =   600

   m(<2, 3, 1>)  =  2^2 * 3^3 * 5^1  = 4 * 27 *   5  =   540

   m(<3, 2, 1>)  =  2^3 * 3^2 * 5^1  = 8 *  9 *   5  =   360

That should afford material to pretend a few hypotheses.

Jon Awbrey

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inquiry e-lab: http://stderr.org/pipermail/inquiry/
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