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 Axiom of choice: Set A needs to be composed of disjunct sets 
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Joined: 31 Mar 2008, 11:54
Posts: 1
Post 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 non-empty 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.

31 Mar 2008, 12:26

Joined: 08 Jun 2011, 14:08
Posts: 1
Post 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 counter-example, 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 member-set. 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.

13 Jun 2011, 18:58
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