Asaf Karagila
I don't have much choice...

It's an almost trivial theorem of cardinal arithmetics in $\ZF$ that given four cardinals, $\frak p,q,r,s$ such that $\frak p<q,\ r<s$ we have $\frak p^r\leq q^s$.

In a recent question on math.SE some user has asked whether or not we always have a strict inequality. Everyone sufficiently familiar with the basics of independence results would know that it is consistent to have $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, in which case taking $\mathfrak{p=r}=\aleph_0,\ \mathfrak{q=s}=\aleph_1$ gives us equality. But it's also trivial to see that we can always pick cardinals whose difference is large enough to keep the inequality true.

So we can always prove that inequality holds for some four cardinals. But what about equality? Can we prove in $\ZFC$ that there are four cardinals for which the exponentiations are equal? Yes. Yes we can, as Andres Caicedo answers the question, taking $\frak p=\beth_\omega$ and $\frak q=\beth_\omega^+$, and $\mathfrak r=\aleph_0,\ \mathfrak s=\aleph_1$ gives us the wanted result. In the comments to his answer, Andres suggested that we try to find a provable example for equality when the axiom of choice fails - for it could very well be that the axiom of choice fails and $\beth_\omega$ is not an $\aleph$, making this equation irrelevant.

I embarked on a few hours of playing with cardinals, and at the end, I have come by the following example. It's easy and nice enough for me to post it here, seeing how I haven't posted anything for quite some time.

Theorem. $\ZF$ proves that there exists four cardinals $\frak p<q,\ r<s$ such that $\frak p^r=q^s$.

Proof. If the axiom of choice holds, then we are done. Assume that the axiom of choice fails, let $N$ be a set which cannot be well-ordered, and let $\newcommand{\fp}{\mathfrak n}\fp$ denote its cardinal. We make the following assumptions:

1. $\fp^\omega=\fp$, otherwise replace $N$ by $N^\omega$. From this assumption we can conclude that: $\fp=\fp+\fp=\fp\cdot\fp$, and from those we can deduce that $\fp^\fp=2^\fp$.
2. If $\kappa=\aleph(\fp)$, then we may assume $\kappa<2^\fp$. If this is not true we can replace $\fp$ by $2^\fp$ (because in that case $\kappa=\aleph(2^\fp)$ as well) or by $2^{2^{\fp}}$ if needed. Since we know that $\aleph(\fp)<2^{2^{2^\fp}}$, one of the three options must have the wanted property.

Note that the properties in the first item are preserved by taking powers, so replacing $\fp$ by its power set (once or twice) would not change the first assumption.

3. Other important properties following from the first property are: $2^\fp=(2^\fp)^\fp=2^\fp+2^\fp=2^\fp\cdot2^\fp.$

We know by a lemma of Tarski that if $\lambda$ is an $\aleph$ and $\mathfrak m$ is a cardinal such that $\frak\lambda+m=\lambda\cdot m$, then $\lambda$ and $\frak m$ are comparable. Because we took $\kappa$ to be incomparable with $\fp$ we know that $\fp+\kappa<\fp\cdot\kappa$.

From the properties of $\fp$ and $\kappa$, we have that $\fp(\fp+\kappa)=\kappa(\fp+\kappa)=(\fp+\kappa)(\fp\cdot\kappa)=\fp\cdot\kappa$.

• Case I: $2^\kappa\nleq2^\fp$. We first observe that $2^\fp<2^{\fp+\kappa}$. Take $\mathfrak p=2^{\fp},\mathfrak q=2^{\fp+\kappa},\mathfrak r=\fp+\kappa, \mathfrak s=\fp\cdot\kappa$. We have that: \begin{align} &(2^\fp)^{\fp+\kappa}=2^{\fp(\fp+\kappa)}=2^{\fp\cdot\kappa}\\ &(2^{\fp+\kappa})^{\fp\cdot\kappa}=2^{(\fp+\kappa)\fp\cdot\kappa}=2^{\fp\cdot\kappa} \end{align} Therefore equality holds, as wanted.
• Case II: $2^\kappa\leq2^\fp$. We note that $2^\fp=2^{\fp+\kappa}$, and now take $\mathfrak p=\kappa,\mathfrak q=2^\fp,\mathfrak r=\fp,\mathfrak s=\fp+\kappa$. Again the inequalities hold, and we calculate: \begin{align} & 2^\fp=2^{\fp+\kappa}\leq\fp^{\fp+\kappa}\leq(2^\fp)^{\fp+\kappa}=2^{\fp\cdot\kappa}=(2^\kappa)^\fp\leq(2^\fp)^\fp=2^\fp\\ & 2^\fp\leq\kappa^\fp\leq(2^\fp)^\fp=2^\fp \end{align} So we have equality again.

We have at least one of the cases true, and our proof is finished. $\square$

(I'd like to thank Andres for suggesting me this very fun exercise in cardinal arithmetics!)

There are 4 comments on this post.

By Harvey Friedman
(Jul 29 2013, 20:37)

Your theorem is an example of an existential sentence about cardinals in the language with only < and exponentiation. Can you determine which sentences in that language are provable in ZF? More generally, expand the set of sentences about cardinals considered, obviously to include addition and multiplication, and perhaps alternating quantifiers.

By
(Jul 29 2013, 21:06 In reply to Harvey Friedman)

Do you mean determine in a syntactical method? Something like $\Sigma_1$-completeness between $\ZF$ (augmented with predicate for cardinals, and their operations along with the axioms for these operations, of course)? That's an interesting question, but I'm not sure how one would approach it.

By
(Nov 09 2014, 22:10)

[…] time ago, Asaf Karagila wrote wonderful post wherein he shows that, even without assuming the axiom of choice one can always find four cardinals […]

By
(Dec 14 2014, 17:37)

[…] Some time ago, Asaf Karagila wrote wonderful post wherein he shows that, even without assuming the axiom of choice one can always find four cardinals (mathfrak{p} lt mathfrak{q}) and (mathfrak{r} lt mathfrak{s}) such that (mathfrak{p}^{mathfrak{r}} = mathfrak{q}^{mathfrak{s}}.) In the comments, Harvey Friedman asks: […]

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