Can it be proved without the axiom of choice that every cardinal is comparable with every finite cardinal?

Question

Can it be proven in ZF, without using the axiom of choice, that every finite set is a universal size comparator, meaning, is comparable with every set in terms of size? And what is the proof?

Answer 1

Yes. For finite cardinals it's obvious from the definition of finite cardinals as the finite ordinals. For infinite sets this is exactly the proof that if $A$ is infinite, then it has finite subsets of arbitrarily large size.

The proof is by induction. If $A_n$ is a subset of $A$ of size $n$, then $A\setminus A_n$ is non-empty, since $A$ is infinite. Choose some element, $a\in A\setminus A_n$ and consider $A_{n+1}=A_n\cup\{a\}$.

You might want to argue that I am making infinitely many choices here. But I'm not. At each stage I only made two arbitrary choices. The first one was choosing some set of size $n$, and then choosing some element outside of it.