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For an infinite cardinal A, then

  1. if $B$ is finite $A^B = A$
  2. If $B$ is infinite and $B \ge A$ then $A^B \ge A^A \ge 2^A > A$
  3. What if $B$ is infinite, but $B < A$, i.e. $A > B \ge \aleph_0$. Is there a relationship between $A^B$ and $A$ in this case ?
Asaf Karagila
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Tom Collinge
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1 Answers1

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We can't quite always say. The continuum hypothesis is equivalent to $\aleph_1^{\aleph_0}=\aleph_1$. But it is not provable, so this equality can hold or fail. But $\aleph_1\leq\aleph_1^{\aleph_0}$ is provable. Therefore, $$\lnot\sf CH\iff\aleph_1<\aleph_1^{\aleph_0}.$$

So the only thing you can really say is that $A\leq A^B$. Nothing much more.

You can also find more intricate problems.

It is always true that $\kappa^{\operatorname{cf}(\kappa)}>\kappa$. So for example, we know that $\aleph_\omega^{\aleph_0}>\aleph_\omega$. But by how much? Is it $\aleph_{\omega+1}$? Is it $\aleph_{\omega+2}$? Is it $2^{\aleph_\omega}$? We can't say, even if we assume that $2^{\aleph_n}<\aleph_\omega$ for all $n<\omega$.

Even worse, what about $\aleph_{\omega_1}^{\aleph_0}$? As far as provability goes, it has no obligation being larger, but it can and it can stay the same (not simultaneously, of course).

But look at $\beth_{\omega_1}$. We can prove that $\beth_{\omega_1}^{\aleph_0}=\beth_{\omega_1}$. So you can't prove exponentiation always grows, and you can prove that it can grow in the case of singular cardinals.

Asaf Karagila
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  • Thanks. If one accepts the continuum hypothesis, is there then a relationship that either $A^B = A$ or $A^B > A$ ? – Tom Collinge Jul 17 '15 at 20:58
  • Well, then you can violate this bond of trust between cardinals on higher levels. Generally you need to assume something like $\sf GCH$ or a specific failure thereof in order to conclude anything particular on cardinals. – Asaf Karagila Jul 17 '15 at 21:01
  • OK: as I understand GHC the infinite cardinals are defined by a complete countably infinite list $\aleph_0, \aleph_1, ......$ where $\aleph_{n+1} = 2^{\aleph_n}$. So if one accepts GCH is there a definite relationship of equality or inferiority between $\aleph_n^{\aleph_m}$ and $\aleph_n$ when $m < n$ ? – Tom Collinge Jul 17 '15 at 21:09
  • Under $\sf GCH$ we can prove the following: $$\kappa^\lambda=\begin{cases}\kappa+ & \lambda<\operatorname{cf}(\kappa)\\lambda^+ &\text{otherwise}\end{cases}$$ And for any ordinal, $\aleph_{\alpha+1}$ is always regular, namely $\operatorname{cf}(\aleph_{\alpha+1})=\aleph_{\alpha+1}$. So in the case of $m,n<\omega$ it immediately implies that $\aleph_n^{\aleph_m}=\aleph_n$ if and only if $m<n$. – Asaf Karagila Jul 17 '15 at 21:13
  • Thanks again: if I understand correctly then if GCH is accepted it follows that if $A > B \ge \aleph_0$ then $A^B = A$ ? – Tom Collinge Jul 17 '15 at 21:18
  • No. You can always prove that $\kappa^{\operatorname{cf}(\kappa)}>\kappa$. So you can always prove that $\aleph_\omega^{\aleph_0}>\aleph_\omega$. – Asaf Karagila Jul 17 '15 at 21:19