Re: Well ordering


José Carlos Santos wrote:
Saurav wrote:

Let alpha be a well orderable infinite set; can we well order 2^alpha
without AC?

As I told you a few days ago, N is well orderable, but without the axiom
of choice you can't prove that 2^N is well-orderable.
Yes, but can you supply a proof that in ZF, this statement requires AC?
Actually, I don't know whether there has been such a proof.

If not, give an example of such an alpha, well orderable,
that within ZF, 2^alpha can not be well ordered.

I can't do that. In ZF you cannot prove that 2^N is well orderable. This
is *not* the same thing as saying that in ZF you can prove that 2^n is
not well orderable.

Best regards,

Jose Carlos Santos

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