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In Uncountable ordinals without power set axiom Francois Dorais explains that without the Power-set Axiom we cannot prove the existence of uncountable ordinals.

I am guess that the power set of an ordinal forces us to go to a higher cardinality, and the axiom of choice forces us to well order that set, therefore we can go to the least ordinal which is uncountable.

Suppose we drop the axiom of choice and add the assertion $\rm{NWP}$: "There is a well ordering of $P(x)$ if and only if $x$ is finite".

In this case we have that $P(\omega)$ (which translates to the power set of any countable ordinal) cannot be well-ordered, and while it can embed all the countable ordinals (as there is a chain which has the order type of the reals) it does not imply the existence of their supremum.

Question: Can we have a model of $\rm{ZF}+\rm{NWP}$ such that the only ordinals are countable? If the answer is yes, can we extend this property to any (regular?) cardinality?

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up vote 12 down vote accepted

No; you can prove the existence of arbitrarily large ordinals in ZF. See the Wikipedia page on Hartogs number and the references therein.

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Silly me, of course there's the Hartogs number! Thanks! –  Asaf Karagila May 11 '11 at 14:43

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