My Math Forum a question about the Axiom of choice

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 August 9th, 2011, 12:01 AM #1 Newbie   Joined: Jun 2011 Posts: 5 Thanks: 0 a question about the Axiom of choice how to proof the following conclusion: for any infinite set S,there exists a bijection f:S--->S*S implies the Axiom of choice. Can you give a proof without the theory of ordinal numbers.
 August 11th, 2011, 04:22 AM #2 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: a question about the Axiom of choice For existential proof, you only need one example. Of all the functions in $f:S\rightarrow S\times S$, the identity function is the easiest. Let $S$ be an indexed set , so $S\neq \emptyset$. It follows $\text{ran }f\neq \emptyset.$ where $\text{ran }f \subseteq S\times S.$ Since $f$ is a function, $\forall i\in S, \langle i,\langle i,i\rangle \rangle \in f$ where $\langle i,f(i)\rangle$ denotes an arbitrary ordered pair. This implies one of the six Axiom of Choice. Conversely, $\forall R \exists f (H\subseteq f^{-1}\; \wedge \; \text{dom }H=\text{dom }f^{-1})$, where $R$ is a relation. Let $y\in \text{dom f^{-1}$. Then $\langle y, H(y)\rangle \in f^{-1}\quad \Rightarrow \langle H(y),y \rangle \in f\quad \Rightarrow f(H(y))=y$. Hence at least an identity function exists. This implies one of the other six Axiom of Choice.

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