# No uncountable ordinals without the axiom of choice?

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?

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

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

It might seems like we can't prove the existence of an uncountable ordinal without the axiom of choice. For example, here's the following incorrect argument:

"Let s be the order type of the set of all countable ordinal numbers, then s must exceed all countable ordinal numbers so it's uncountable"

The problem with that argument is that since there is no set of all ordinal numbers, that argument assumes there's an uncountable ordinal number to construct the set of all countable ordinal numbers to define the order type of.

Even so, it can be proven anyway with the axiom of a power set. It can be shown that there are at least as many real numbers as there are well ordering relations on the set of positive integers. The set of all such well ordering relations can be grouped in such a way that 2 such well ordering relations are in the same group if and only if they have the same order type. From that, we can construct the set of all order types of any such group. That's the set of all countable ordinal numbers. The order type of that set with the natural order relation can be shown to be the first uncountable ordinal number.

• You did see that this is five years old now, right? – Asaf Karagila May 13 '16 at 3:34