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Martin Sleziak
Martin Sleziak
7:00 AM
For cardinals such that $a,b\ge\aleph_0$ we have $a+b=ab=\max\{a,b\}$ (assuming AC).
But for $a=\aleph_0$ the fact about addition can be shown without AC. (It is basically Hilbert's hotel argument.)
I.e., we have $\aleph_0\le b$ $\implies$ $\aleph_0+b-b$.
It is natural to ask whether something similar holds for multiplications. It turns out that this is not provable in ZF alone.
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Q: Does $\aleph_0\cdot\kappa=\kappa$ for every $\kappa\ge\aleph_0$ hold in ZF?

Martin SleziakIt is easy to show that for any (Dedekind) infinite cardinal $\kappa$ we have $\aleph_0+\kappa=\kappa$. Definition of an infinite cardinal is a cardinal such that $\aleph_0\le\kappa$. (I believe this is usually called Dedekind infinite.) We can use basically the same "Hilbert hotel argument" as...

set-theory cardinals axiom-of-choice
This shows how forgetful I am. I thought about this problem a bit. In only after deciding to search for the question on math.SE, I thought: "Wait a minute! Didn't I ask exactly this question on this site some time ago?"
 

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