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It 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 in the proof of $\aleph_0+\aleph_0=\aleph_0$. This proof does not use Axiom of Choice.

With the use of AC we can show $\aleph_0\cdot\kappa=\kappa$.

We have $\kappa\le\aleph_0\cdot\kappa \le \kappa\cdot\kappa = \kappa$. Then we can use Cantor-Bernstein theorem. In the equality $\kappa\cdot\kappa = \kappa$ we are using Axiom of Choice. (See here, here or here.)

So my question is:

Can $\kappa\cdot\aleph_0=\kappa$ for every $\kappa\ge\aleph_0$ be shown in ZF (i.e., without using Axiom of Choice)?


Disclaimer: I strongly suspect that the answer to this question can be very probably found in some of the other answers on this site or can be derived as a consequence of some result related to choice mentioned in other answers. But I did not find a questions asking explicitly this. I thought it might be useful to have such question as a reference. (If not for other reason, this might help people searching for the answer to this question.)

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    $\begingroup$ We can't show in ZF that $\aleph_0+\kappa=\kappa$. If we take $\kappa$ to be so called infinite Dedekind-finite cardinality, then we don't have $\aleph_0\leq\kappa$, so mentioned equality cannot hold. Same goes for multiplication. $\endgroup$ – Wojowu Jan 4 '15 at 8:49
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    $\begingroup$ @Wojowu I should have made it clearer, but I took $\aleph_0\le\kappa$ for the definition of infinite cardinal. (I should probably wrote Dedekind infinite to make that clear.) I have edited the post. Thanks for pointing this out. $\endgroup$ – Martin Sleziak Jan 4 '15 at 8:55
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No, take any infinite Dedekind finite set $a$ and consider $X=a\cup\omega$. Then $X\times\omega$ has a strictly larger cardinality than $X$.

To see that simply note that it embeds two copies of $a$, whereas $X$ can only embed one.


More generally, we can show that $a+b=a$ if and only if $a+b\cdot\aleph_0=a$ without using the axiom of choice. This means that $a\cdot\aleph_0=a$ implies that $a+a=a$, something which is not true, even if we assume arbitrarily large $\sf DC_\kappa$ holds.

(The above can be concluded since the existence of $\kappa^+$-amorphous sets is consistent with $\sf DC_\kappa$.)

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