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| ``Re: Elegant proof (AC ==> Zorn)''
by smw on 2006-02-27 10:28:40 |
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| That is correct. I am thinking that perhaps I should not have stated "and this is closer to Jech's proof." I really thought that what I was saying was true. Upon reconsideration I believe that I was not as informed as I thought I was.
Of course, I was always aware that Jech was using the power set. I didn't really understand why until just now as I am writing this. Jech's proof is influenced by the way in which he formulates the axiom of choice. I just checked the "canonical" formulation of AC here on PM and it agrees completely with Jech's formulation. So using the power set as the domain of the choice function *is* the natural thing to do here.
I thought that my "revision" was not only simpler than the PM proof, but also Jechs. In reality, it is not simpler in any objective sense. It seems simpler to me because I walk around with a different "canonical" version of the axiom of choice in my head, namely:
``If {A_x}_{x in X} is a collection of nonempty sets, indexed by some set X, then there is a choice function f: X ---> U A_x such that f(x) is in A_x for all x in X.''
More succinctly, ``the cartesian product of two nonempty sets is nonemtpty.'' I guess it is this version of the axiom of choice that I have found to be the most practical, so it sort of stuck. |
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