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?

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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.

Asaf Karagila

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