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Re: Why not Hausdorff's ordered pairs
Posted:
Feb 20, 2012 6:11 PM
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Are you waiting for me to answer your question? But, you know the answer anyway, don't you?
The definition is not sufficient because there is no difference between {{1,a}, {2,b}} and {{1,a}, {b,2}}. That makes <1,1> = {{1,1}, {2,1}} = {{1,2}, {2,1} as a duplicate, {1,1}} = <2,1> as a possibility, which is not what you want. You don't want a definition or an axiom to prove what you don't want.
The question is, what makes order? Inaccessibility of 2 unless through 1 (hierarchy-based, n to sort)? Dependency of 2 on 1 (counting-based, n to sort)? Or, the existence of a binary relation < to order them (comparison-based, n*log(n) to sort)? There are many, many ways to define order. Which gives you everything you want? None of the above?
The comparison-based approach works on P(N) and above though. Counting fails with uncountability, as we all know.
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