1. Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$, where $\mathfrak{ p,q,r,s}$ are cardinal numbers?
  2. Is it possible to prove in $\mathsf{ZFC}$ that there is a counterexample?
  • I don't think ZFC can prove the statement of interest. Here's my logic. If ZFC is consistent, then it has a model in which there exist cardinal numbers $\mathfrak{r}$ and $\mathfrak{s}$ such that $\mathfrak{r} < \mathfrak{s}$, but $2^\mathfrak{r} = 2^\mathfrak{s}$. Lets work in such a model, and fix any such $\mathfrak{r},\mathfrak{s}.$ Then these cardinals are necessarily infinite, so $2^\mathfrak{s}=3^\mathfrak{s}.$ So take $\mathfrak{p}=2$ and $\mathfrak{q}=3$. – goblin May 26 '13 at 15:29
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    I think that one can use $\beth_\omega$ as a basis for a counterexample in $\sf ZFC$. Let $\kappa=\beth_\omega$ and $\lambda=\kappa^{\aleph_0}$, then $\lambda>\kappa$ and $\lambda^{\aleph_1}$ should be equal to $\kappa^{\aleph_0}=\lambda$. But I'm not 100% certain about that, so I'm leaving this as a comment for now. – Asaf Karagila May 26 '13 at 16:52
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    On second thought, I think it might be consistent that $\operatorname{cf}(\lambda)=\aleph_1$, in which case the above example is false... – Asaf Karagila May 26 '13 at 16:59
up vote 16 down vote accepted

The inequality does not hold in general. First, note that $2^\kappa=\kappa^\kappa$ for any infinite $\kappa$. Second, it is consistent (using the technique of forcing) that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, even though $\aleph_0<\aleph_1$. This gives us that it is consistent that $\aleph_0<\aleph_1$ and yet $\aleph_0^{\aleph_0}=\aleph_1^{\aleph_1}$.

For a (combinatorially more involved) $\mathsf{ZFC}$ example, $\beth_\omega^{\aleph_0}=\beth_\omega^{\aleph_1}$, so $\beth_{\omega}^{\aleph_0}=(\beth_\omega^+)^{\aleph_1}$. (For a similar computation, see Jech's exercise 5.18, here or here.)

  • You're better off writing, "The inequality does not hold in all models of ZFC," in my opinion. Otherwise, the first sentence seems to be suggesting that, in every model of ZFC, there exists a counterexample. – goblin May 26 '13 at 16:23
  • Ah, so my first comment on the question was almost correct! Thanks! – Asaf Karagila May 26 '13 at 17:31
  • @AsafKaragila How about a $\mathsf{ZF}$ example? Maybe starting with some non-well-orderable $X$ such that $\aleph(X)$ injects in $2^X$, and comparing $X^X$ with $(X+\aleph(X))^{X+\aleph(X)}$. – Andrés E. Caicedo May 26 '13 at 17:34
  • @Andres: That sounds like a reasonable possibility. I'll work the details after I get home. Your suggestion somewhat relates to an MO question of mine about $2^X=X^X$ and the axiom of choice. – Asaf Karagila May 26 '13 at 17:37
  • @AsafKaragila Yes, this wouldn't be an arbitrary $X$. You want at least $X+X\sim X$, maybe even $X\times X\sim X$, and $\aleph(X)\le_* X$, or somesuch. – Andrés E. Caicedo May 26 '13 at 17:39

Andres Caicedo asked whether or not we can prove in $\sf ZF$ that there exists a counterexample (i.e. a situation where inequality holds in the assumption, but the exponents are equal).

In $\sf ZFC$ we know that there exists such counterexample, so let us assume $\sf ZF+\lnot AC$. Let $P$ be a set which cannot be well-ordered, and let $\newcommand{\fp}{\mathfrak p}\fp$ denote its cardinal. We make the following assumptions:

  1. $\fp^\omega=\fp$, otherwise replace $P$ by $P^\omega$. 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 $\kappa$ is the least ordinal not less or equal than $\fp$, then $\kappa<2^\fp$. If this is not true we can replace $\fp$ by $2^\fp$ (because $\kappa$ is the least ordinal with that property for $2^\fp$ as well) or by $2^{2^{\fp}}$ if needed. Since we know that $\kappa<2^{2^{2^\fp}}$, one of the three options must have the wanted property.

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

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 $\fp+\kappa<\fp\cdot\kappa$.

Note that $\fp(\fp+\kappa)=\kappa(\fp+\kappa)=(\fp+\kappa)(\fp\cdot\kappa)=\fp\cdot\kappa$ by the properties of $\fp$ and $\kappa$.

  • Case I: $2^\kappa\nleq2^\fp$.

    We observe that $2^\fp<2^{\fp+\kappa}$, and now we calculate: $$\begin{align} &(2^\fp)^{\fp+\kappa}=2^{\fp(\fp+\kappa)}=2^{\fp\cdot\kappa}&\tag{1}\\ &(2^{\fp+\kappa})^{\fp\cdot\kappa}=2^{(\fp+\kappa)\fp\cdot\kappa}=2^{\fp\cdot\kappa}&\tag{2} \end{align}$$ And it is not hard to see that the inequalities are satisfied, but the exponentiation ends up equal.

  • Case II: $2^\kappa\leq2^\fp$.

    We note that $2^\fp=2^{\fp+\kappa}$, and we make the following calculations: $$\begin{align} & 2^\fp=2^{\fp+\kappa}\leq\fp^{\fp+\kappa}\leq(2^\fp)^{\fp+\kappa}=2^{\fp\cdot\kappa}=(2^\kappa)^\fp\leq(2^\fp)^\fp=2^\fp&\tag{3}\\ & 2^\fp\leq\kappa^\fp\leq(2^\fp)^\fp=2^\fp &\tag{4} \end{align}$$ And in this case we see that $\kappa<2^\fp$ and that $\fp<\fp+\kappa$, but the exponentiation is again equal.

  • Very nice!${}{}$ – Andrés E. Caicedo May 27 '13 at 0:44
  • Thank you! And thank you for pointing me to this direction, I haven't been very productive recently, and this just might be the trigger for a new wave of productivity... I hope! – Asaf Karagila May 27 '13 at 0:48
  • 2
    I'm intrigued by the downvote. – Asaf Karagila Jul 22 '13 at 19:37

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