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Axiom of choice: Set A needs to be composed of disjunct sets http://www.roadtoreality.info/viewtopic.php?f=20&t=35 
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Author:  Sino [ 31 Mar 2008, 12:26 ] 
Post subject:  Axiom of choice: Set A needs to be composed of disjunct sets 
I've just started reading and came across the Axiom of Choice in the 1st chapter that was briefly mentioned there. Since I wanted to know, what it states, I went immediatly to 16.3 ( and figure 16.4 ). The description at the figure says "for any set A, all whose members are nonempty sets, there exists a set B, which contains exactly one element from each of the sets belonging to A.". And in the text it's stated in a similar way. After a minute thinking I came up with the set A={{1},{2},{1,2}}, for which no such set B exists. So I checked wikipedia for the Axiom of Choice, which requires all sets in A to be pairwise disjunkt. The members in my example set A aren't pairwise disjunkt of course and that's why it fails. For a mathematician that inaccuracy is probably trivial and he doesn't even notice it, because he knows the Axiom of Choice already, but for a normal reader like me, it might be a bit irritating at first. So I think this should be corrected. It's on page 366 paperpack edition (vintage). Anyway, like I've said I've just started with the book and think it's great so far. 
Author:  Wim Nobel [ 13 Jun 2011, 18:58 ] 
Post subject:  Re: Axiom of choice: Set A needs to be composed of disjunct 
I just discovered this forum and already found this post to which I wish to reply. This is, because I do not agree that there's an inaccuracy here. The Axiom of Choice does not require that the sets which are elements of set A are pairwise disjoint. I also checked this in Wikipedia, but found no assertion of that kind there (it does, however, mention a variant in which it is required). Also, the given case is not a proper counterexample, because the set B = {1,2} meets the requirements of the Axiom of Choice. Note that it is not required that all chosen elements are different. Suppose we choose the element 1 from the third memberset. Then the enumeration of all chosen elements is 1, 2, 1. The set containing all these elements is denoted {1,2} because mentioning an element twice does not yield a different set. 
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