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.


Are there any results known about this? To be more specific, let me ask the following question:



> 
> 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.



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


