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  1. Will ordinals be well-ordered without AC? This seems to be obviously true, as they are by definition well-ordered.

  2. Why would we then need the axiom of choice. We can just form a bijection function to ordinals, allowing all sets to be well-ordered.

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How would you form your bijection function? –  Ben Millwood Sep 16 '12 at 11:49
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up vote 7 down vote accepted

Ordinals are defined without the axiom of choice. Ordinals are simply sets such that $\in$ induces a transitive and well-ordered relation on them. Therefore the ordinals are always well-ordered.

The problem is that without the axiom of choice you cannot always put things in bijection with ordinals. For example you cannot define a bijection between the real numbers and an ordinal without the axiom of choice.

Assuming the axiom of choice if two models of set theory have the same sets of ordinals then they are isomorphic. So in some sense the ordinals really do form the backbone of the universe if choice is assumed. However without the axiom of choice this is not true any more. You can find two models which have the same ordinals and same sets of ordinals, but not the same sets of sets of ordinals... this can be extended indefinitely too (sets of sets of sets ... of ordinals).

Remember that not all sets are ordinals, or sets of ordinals. At least without the axiom of choice.

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  1. Yes, ordinals require very littel axiomatic framework. For example, one does not need the foundation axiom to prove that ordinals are well-founded. Also, they are almost automatically well-ordered.

  2. The statement "For all sets $X$ there is an ordinal $\alpha$ and a bijective $f\colon\alpha\to X$" is not weaker than the Axiom of Choice: It both shows that and follows from the statement that all sets can be well-ordered, which is well-know to be equivalent to AC (on the basis of ZF).

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(1) depends a bit on the definitions one works with. Some possible definitions of "ordinal" work only in the presence of Foundation. –  Henning Makholm Sep 16 '12 at 12:23
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