Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, ...
Given the number of results that are independent of ZF. It seems that once you've found a proof of a theorem that uses the axiom of choice, the odds are that it will be independent of ZF. So my ...
Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice?
Let X be an infinite set. Is it possible to show the existence of a countably infinite subset of X without using the Axiom of Choice?
I recently saw the proof of the independence of ZF (with allowance for multiple empty sets) and AC. The proof constructed the model based on a set theory generated by infinitely many empty sets and ...