I know that order type $\omega + 1$ can be well-ordered, but I am not sure if real numbers on $[0,1]$ can be well-ordered.
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8Assuming the axiom of choice, every set can be well ordered, and vice-versa- – Pedro Nov 10 '14 at 20:24
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4Yes, they can. No, no "explicit" well-ordering can be exhibited. The issue is that the statement that the reals are well-orderable is a theorem of the standard system of axioms for set theory (that includes the axiom of choice), but it is not provable if we do not assume some (reasonably strong) sort of choice-like principle. – Andrés E. Caicedo Nov 10 '14 at 20:25
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As a supplement to @Andres's comment, it is consistent with ZF that $\Bbb R$ cannot be well ordered. – Nov 10 '14 at 20:37
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Of course they can.
Start numbering them by ordinal numbers $r_0, r_1, r_2, \ldots, r_\omega, r_{\omega+1}, \ldots$ and sooner or later, all reals in $[0,1]$ will be depleted because there are more ordinals than the size of any set.
user2345215
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Are you claiming that any set can be well ordered? Your argument doesn't make sense to me. If it's correct, then you're using the axiom of choice somewhere, but it's not clear where. – Nov 10 '14 at 20:31
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1@MikeMiller I'm also using the axiom of replacement in the final step, beware. – user2345215 Nov 10 '14 at 20:34
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4Downvoter, you mad? Go back to peano arithmetic if you don't like ZFC. – user2345215 Nov 10 '14 at 20:36