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.)

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $\sf DC_\kappa$.)

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.