Talk:Condition for Set Union Equivalent to Associated Cardinal Number

What I don't understand about this theorem is that it is a fact by definition of set equivalence and cardinality that $S \cup T \sim \left|{S \cup T}\right|$. Similarly it is also a fact that $S \sim \left|{S}\right|$ and $T \sim \left|{T}\right|$.

So this theorem boils down to the equivalent of (using basic arithmetical conventions) $1 + 1 = 2 \iff 2 + 2 = 4 \land 3 + 3 = 6$. Each statement is true independently of each other. The same applies to the theorem proved here.

What am I missing? --prime mover (talk) 20:10, 1 September 2012 (UTC)

Cardinal Numbers do not work the same way. Without the axiom of choice, we cannot prove that $S \sim |S|$. If we could prove this, then we could prove that $S$ is well-ordered. The fact that $S \sim |S|$ is only true if we accept the axiom of choice. --Andrew Salmon (talk) 21:10, 1 September 2012 (UTC)

Oh okay, so I missed the bit where it says $S \sim |S| \iff $ AOC. Can a link to that page be added to this one? --prime mover (talk) 22:14, 1 September 2012 (UTC)