The Road to Reality

Axiom of choice: Set A needs to be composed of disjunct sets
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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 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.

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 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.

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