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Old 05-30-2003, 07:34 AM   #31
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Default Axiom of Choice, delusion or grandeur?

Thanks for your clear response.

Quote:
Originally posted by Bookman
In real-world examples, several such functions present themselves: let f(x) yield the cheapest, the fattest, the eldest and so on.
Isn't this merely the intersection between the set of x and the subset that has the properties for which f(x) is true? If the property we're looking for is "eldest" and that property is such that it can pertain to only one member of the set then yes, we can compare each set member to determine which one it is.

My continued observation is that the "set of sheep" (as opposed to the real sheep) comprises members that have the quality "sheepness" within the space/time domain in which the real sheep exist. The "set of sheep" contains no information as to other qualities - a function to determine "eldest" is applied to the real sheep.

Note: If the function(eldest) is applied to the "set of sheep" (as opposed to the real sheep) it seems clear that an anomalous result will obtain since one is comparing apples and oranges, so to speak. With apples=~oranges its inevitable that passing a function over the apples to determine the orangiest could result in the mistaken result that apples=~apples.

Cheers, John
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Old 05-30-2003, 09:24 AM   #32
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I'm beginning to understand the point that you're making. It seems to me that strictly speaking your problem is not with the Axiom of Choice but rather with choice functions in general.

I'll extend my earlier response to hopefully deal with your objection.

The function eldest might be expressed symbolically as:
Eldest(A) = {x e A | age(x) >= age(a') V a' e A}
(taking the lowercase e to act as the union operator as I can't find that character in unicode...)

"The eldest of A is the set of all x in A such that the age of x is greater than the age of a' for each a' in A"

The result set S is then expressed as:
Eldest(A) ^ Eldest(B) ^ Eldest(C)
(...and taking the carat to act as the union operator as I can't find that character in unicode either.)

As you can see, the Eldest function (operating on a set) is defined in terms of the age operation (operating on a sheep).

Quote:
Isn't this merely the intersection between the set of x and the subset that has the properties for which f(x) is true?
Yes. It is trivially true that the results of the choice function is the intersection of itself with the domain of sets.

Bookman
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