Talk:Diaconescu's theorem

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In the second formula, I changed "{0} to {1} if not P". Assuming U refers to the first set and V to the second, "not P" implies "x = 1" which rules out "V = {0}". MichaelShoemaker (talk) 21:22, 17 July 2009 (UTC)[reply]

I changed it back. On rereading, my interpretation was not correct. MichaelShoemaker (talk) 21:25, 17 July 2009 (UTC)[reply]

This proof is a strawman argument

IMHO, the correct intepretation of the AC is: given a family of sets $S_{i},i\in I,$ there exists a function $f$ such that $\forall i:f(i)\in S_{i}$ . It is a much more reasonable interpretation: whenever we have choice involved in practice, we don't just have sets, we have sets in certain roles, here represented by the indices.

This way, if we took $I=\{0,1\},$ $S_{0}=U,$ $S_{1}=V,$ then $S_{0}=S_{1}$ would not imply $f(0)=f(1)$ , i.e. we could still make different choices, and the truth of $P$ would not be recoverable.

The proof weasels in the otherwise completely unwarranted $U=V\implies f(U)=f(V)$ step by having the choice function act on the values and not on the roles.