# How much choice does a linear or well-order on cardinals imply?

It is well-known that if the natural (partial) order on the class of cardinal numbers is a linear order, then it is in fact a well-order and the axiom of choice holds. I was, however, interested in how much choice we can recover given some linear ordering, or better — a well-ordering — of the class of cardinals.

I couldn't figure out by myself any results, but I would imagine it at least implies that there are no amorphous sets.

Is the axiom of countable choice implied by the existence of a linear ordering on the class of all cardinals? How about the existence of a well-order on this class?

Thanks in advance for all the feedback.

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No.

In Cohen's first model, as constructed over $L$, you have a uniform linear ordering of the entire model. However the axiom of countable choice fails: there is a Dedekind-finite set of reals.

To see that the first claim holds, note that the model is $L(A)$ where $A$ is the of generic Cohen reals. Therefore there is a surjection from $[A]^{<\omega}\times\sf Ord$ onto $L(A)$, and that defines a linear ordering on the entire universe. In particular on the class of cardinals.

(Remark: I am not claiming that the order extends the order of the cardinals, but it does order the class of cardinals (which is $\omega$, the $\aleph$ numbers and the Scott cardinals) in a linear order. Whether or not there is an order extension of the usual order to a linear order, I cannot say. I suspect this is the case in Cohen's model, though. Note it cannot be extended to a well-ordering since there are infinite decreasing chains.)

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That seems to do the question with linear order. I suppose this can't be modified to answer the question of well-ordering of cardinals implying countable choice. – Wojowu Jan 9 at 20:21
Yeah, I don't know about that. I'd be surprised if you can well-order something like the cardinals; but not surprised enough to be shocked. – Asaf Karagila Jan 9 at 20:23
@Joel: I think this touches a deeper question. What are cardinals? For me they are the aleph numbers and Scott cardinals. These are all sets, and they get linearly ordered with the rest of the universe. – Asaf Karagila Jan 10 at 4:56
Ah, I understand now. The point should be that if you can linearly order the universe at all, then you define a new linear order, which does respect equinumerosity. Namely, put the Scott cardinals in order (since as you say they are sets, definable representatives for each equinumerosity class), and then order within each equinumerosity class by the original order. In this way, you arrive at the kind of linear order I had wanted. (And in particular, this doesn't imply choice.) – Joel David Hamkins Jan 10 at 5:11
@AsafKaragila I don't require "$a\leq b$ implies $a$ has cardinality at most $b$". I only require "$a\leq b$ and $b\leq a$ implies $a$ has the same cardinality as $b$". So you did interpret my question correctly. – Wojowu Jan 10 at 13:44