[lbo-talk] Re: Ambigious Doug

Chuck Grimes cgrimes at rawbw.com
Sat Jun 28 09:48:46 PDT 2003


``There is indeed a difficulty about having a concept of 'everything', for we ordinarily conceive of something with, so to say, a boundary around it: this is a sheep and not a giraffe. But everything is bounded by nothing...'' CG Estabrook

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Sorry I didn't read this post earlier. Virtually identical arguments can be formulated with the theory of sets. So that the discussion about everything and nothing takes a mathematical form. To spare anyone of the agony of doing the math, the answer is that Nothing contains Everything!...and has to be an axiom about anything.

It is the major principle of set theory and has the German name, Aussonderungsaxiom. I think it is also known as the axiom of existence or the axiom of comprehension.

What follows is adapted from Paul Halmos, Naive Theory of Sets, Van Nostrand, NY, 1967, p6.

Axiom of Specification. To every set A and to every condition S(x) there corresponds a set B whose elements are exactly those elements x of A for which S(x) holds.

The above axiom is given symbolically as:

B = {x e A: S(x)}.

In other words there is a set B which equals both the elements x that belong to A and some function of the elements x, S(x). For example, not (x e x) = S(x).

The proof is simple but tough to understand.

Can it be that B belongs to A? There are two possibilities. First. If B belongs to A, then either B belongs to B or else B does not belong to B. If B belongs to B, then the assumption that B belongs to A, means B does not belong to B. This is a contradiction. Second. If B does not belong to B, then the assumption, B belongs to A means B belongs to B, which is also a contradiction. It is impossible that B belongs to A.

The connection with the argument of existence is that we take B to be nothing then show it does not belong to anything and then propose it equals everything, ie. x e A and S(x).

Chuck Grimes



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