What is the cofinality of $\kappa^{<\kappa}$?

Question

I am not sure if this has been answered before, but I did not find anything.

Definition $\kappa^{<\kappa}=\sup\{\kappa^\lambda\mid\lambda<\kappa\}$.

Assume $\kappa$ to be limit and singular. What do we know about the cofinality of $\kappa^{<\kappa}$?

Answer 1

Suppose $\kappa$ is a singular cardinal. To analyze the size of $\kappa^{<\kappa}$, there are two cases to consider: either

1. the function $\rho\mapsto\kappa^\rho$ eventually stabilizes below $\kappa$, that is, there is a $\tau<\kappa$ such that for all $\rho$ with $\tau\le\rho<\kappa$ we have $\kappa^\rho=\kappa^\tau$, or else
2. case 1. fails, that is, the function $\rho\mapsto\kappa^\rho$ is strictly increasing on a cofinal set of cardinals below $\kappa$.

In case 2., obviously $\kappa^{<\kappa}$ has cofinality $\mathrm{cf}(\kappa)$.

Let's now consider case 1. We argue that $\kappa^{<\kappa}=2^\kappa$ in this case. It follows that its cofinality is strictly larger than $\kappa$, and we cannot really say more.

To prove the equality, first we prove the lemma that, in general, $$2^\lambda=(2^{<\lambda})^{\mathrm{cf}(\lambda)}.$$ This is obvious if $\lambda$ is a successor, so it suffices to consider the case where $\lambda$ is limit. Consider a strictly increasing sequence $(\lambda_\alpha)_{\alpha<\mathrm{cf}(\lambda)}$ cofinal in $\lambda$. We have $$2^\lambda=2^{\sum_\alpha \lambda_\alpha}=\prod_\alpha 2^{\lambda_\alpha}\le(2^{<\lambda})^{\mathrm{cf}(\lambda)},$$ and the other inequality is clear.

Returning to case 1., there are again two options: (a) If for some $\rho<\kappa$ we have $2^\rho\ge\kappa$ then $\kappa^\rho=2^\rho$. Since we are in case 1., it follows that in this situation the function $\rho\mapsto2^\rho$ eventually stabilizes below $\kappa$, and of course $\kappa^{<\kappa}$ is also equal to this value. In this case, $2^\kappa$ is also equal to this common value, by the lemma, and we are done.

(b) Otherwise, $\rho<\kappa$ implies $2^\rho<\kappa$, that is, $\kappa$ is a strong limit singular cardinal. In this case, $2^\kappa=\kappa^{\mathrm{cf}(\kappa)}$, by the lemma. Since $\kappa^\kappa=2^\kappa$, it follows that $\kappa^{<\kappa}=\kappa^{\mathrm{cf}(\kappa)}$ and again we are done.

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Let me conclude by pointing out that all cases mentioned in the analysis above are possible:

Case 2 is realized for $\kappa=\aleph_\omega$ if $2^{\aleph_n}=\aleph_{\omega+n+1}$ for all $n<\omega$. This can be easily achieved starting with a model of $\mathsf{GCH}$ and using Easton forcing. In this case we have $\aleph_\omega^{\aleph_n}=2^{\aleph_n}$ for all $n$ (and $\aleph_\omega^{<\aleph_\omega}=\aleph_{\omega+\omega}$ has cofinality $\omega=\mathrm{cf}(\aleph_\omega)$).

One may be curious whether this can be improved to have $\aleph_\omega$ strong limit and $\aleph_\omega^{\aleph_0}<\aleph_\omega^{\aleph_1}<\dots\,$. It turns out this is not possible, as proved independently by Hajnal and Shelah. Indeed, if $\kappa$ and $\lambda$ are infinite cardinals and $2^{<\lambda}\le\kappa$, then $\{\kappa^\rho\mid \rho<\lambda\}$ is finite. I learned this as Theorem 1.5 in

MR0991617. Monk, J. Donald. Appendix on set theory. In Handbook of Boolean algebras, Vol. 3, 1213–1233, North-Holland, Amsterdam, 1989.

(I looked at this appendix just now, to find the precise reference, and it seems to me that it also contains enough information to reconstruct my answer.)

Case 1.(a) is possible by starting with a model of $\mathsf{GCH}$ and forcing $2^{\aleph_0}=2^{\aleph_\omega}=\aleph_{\omega+1}$, simply by adding Cohen reals. In this case we have $\aleph_\omega^{<\aleph_\omega}=\aleph_\omega^{\aleph_n}=\aleph_{\omega+1}$ for all $n$ (a regular cardinal).

We could have just as easily added many more Cohen reals and get, say, $2^{\aleph_0}=2^{\aleph_\omega}=\aleph_{\omega_{\omega+3}}$, which now makes $\aleph_\omega^{<\aleph_\omega}$ a singular cardinal of cofinality $\aleph_{\omega+3}$.

Case 1.(b) holds under $\mathsf{GCH}$. Under appropriate large cardinal assumptions, we can make life more interesting by forcing a violation of the singular cardinal hypothesis at $\aleph_\omega$ (while keeping $\aleph_\omega$ strong limit). There are now some limitations, coming from Shelah's pcf theory but, for instance, we can have $\aleph_\omega$ strong limit and $2^{\aleph_\omega}=\aleph_\omega^{<\aleph_\omega}=\aleph_{\alpha+1}$ for any infinite countable ordinal $\alpha$, as shown in

MR0823777 (87b:03114). Shelah, Saharon. The singular cardinals problem: independence results. In Surveys in set theory, 116–134, London Math. Soc. Lecture Note Ser., 87, Cambridge Univ. Press, Cambridge, 1983.

Pcf theory tells us we cannot get $\alpha$ above to be $\aleph_4$ or higher, but the picture here is still far from clear.