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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. — Preceding unsigned comment added by 78.83.51.215 (talk) 23:06, 18 June 2011 (UTC)[reply]

Taken back. We can take the family to be

\{U_{U},V_{V}\}

, in which case

U=V\implies f(U)=f(V)

would still be relevant. 78.83.51.215 (talk) 23:39, 18 June 2011 (UTC)[reply]

The proof still stinks though. We don't need AC to choose from two sets (or from any finite collection of sets, for that matter,) so, by this proof, excluded middle seems to be a tautology. 78.83.51.215 (talk) 00:00, 19 June 2011 (UTC)[reply]

Revision as of 23:40, 18 June 2011 edit 78.83.51.215 (talk) →‎This proof is a strawman argument ← Previous edit		Revision as of 00:00, 19 June 2011 edit undo 78.83.51.215 (talk) →‎This proof is a strawman argument Next edit →
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	: Taken back. We can take the family to be <math>\{ U_U, V_V \}</math>, in which case <math>U = V \implies f(U) = f(V)</math> would still be relevant. [[Special:Contributions/78.83.51.215\|78.83.51.215]] ([[User talk:78.83.51.215\|talk]]) 23:39, 18 June 2011 (UTC)		: Taken back. We can take the family to be <math>\{ U_U, V_V \}</math>, in which case <math>U = V \implies f(U) = f(V)</math> would still be relevant. [[Special:Contributions/78.83.51.215\|78.83.51.215]] ([[User talk:78.83.51.215\|talk]]) 23:39, 18 June 2011 (UTC)

			: The proof still stinks though. We don't need AC to choose from two sets (or from any finite collection of sets, for that matter,) so, by this proof, excluded middle seems to be a tautology. [[Special:Contributions/78.83.51.215\|78.83.51.215]] ([[User talk:78.83.51.215\|talk]]) 00:00, 19 June 2011 (UTC)