# Question about $\kappa^{<\kappa} = \kappa$

Set theory noob here...

So I found a few results stating that if a cardinal $$\kappa$$ fulfills some property X, then $$\kappa^{<\kappa} = \kappa$$ (e.g. X is '$$\kappa$$ is strongly inaccessible', see If $\kappa$ is strongly inaccessible, then $\kappa^{<\kappa} = \kappa$ ).

My question concerns the other implication direction. If we assume that $$\kappa^{<\kappa} = \kappa$$ for some $$\kappa$$, what can we say about the "size" of $$\kappa$$? Is there an informal way to think about such cardinals? (maybe assume that $$\kappa$$ is uncountable)

• Isn’t it equivalent to asking that the cofinality of $\kappa$ be $\kappa$? Oct 30, 2022 at 8:43
• @Aphelli In that case, I'd be wondering what exactly one would like to stress when using the above form instead of just stating that the cardinal should be regular. Maybe that a cardinal can be used to count the set of all maps $\lambda \rightarrow \kappa$ where $\lambda < \kappa$? Maybe the question is too wide though. (Trying to build up some intuition here) Oct 30, 2022 at 8:54
• That’s possible (I’m not sure about the notation either), but even so: if $\kappa$ has cofinality less than $\kappa$, then $\kappa^{< \kappa}$ is greater than $\kappa$. If $\kappa$ is regular, then for all nonzero $\lambda <\kappa$, $\kappa^{\lambda} =\kappa$. Then taking the union over all $\lambda <\kappa$, you get $\kappa^{< \kappa} \leq \kappa^2=\kappa$. Oct 30, 2022 at 12:25
• @Aphelli "If $\kappa$ is regular, then for all nonzero $\lambda<\kappa$, $\kappa^\lambda = \kappa$." This is wrong. For example, if CH fails, then $\aleph_1^{\aleph_0} \geq 2^{\aleph_0} > \aleph_1$, so $\aleph_1^{\aleph_0} \neq \aleph_1$, but $\aleph_1$ is regular. Oct 30, 2022 at 14:22
• @AlexKruckman: very good point (and very serious mistake on my part), thank you for bringing this up! I was sure that if $\lambda$ was less that the cofinality of $\kappa$ (with $\lambda$ and $\kappa$ infinite, say), then $\kappa^{\lambda}=\kappa$. I guess the correct statement would be $max(\kappa,2^{\lambda})$ instead… and then the “strongly inaccessible” part becomes much clearer… Oct 30, 2022 at 15:21

The following are equivalent for an infinite cardinal $$\kappa$$:

1. $$\kappa^{<\kappa} = \kappa$$
2. $$\kappa$$ is regular and for all $$\lambda < \kappa$$, $$2^\lambda \leq \kappa$$.
3. (a) $$\kappa = \aleph_0$$ or (b) $$\kappa$$ is (weakly) inaccessible with $$2^\lambda\leq \kappa$$ for all $$\lambda < \kappa$$, or (c) $$\kappa = \mu^+$$ and GCH holds at $$\mu$$: $$2^\mu = \kappa$$.

Proof: $$1\implies 2$$: Assume $$\kappa^{<\kappa} = \kappa$$. Suppose for contradiction that $$\kappa$$ is singular. Then $$\kappa^{<\kappa} \geq \kappa^{\mathrm{cf}(\kappa)} > \kappa$$. Now suppose for contradiction that there exists $$\lambda<\kappa$$ such that $$2^\lambda > \kappa$$. Then $$\kappa^{<\kappa}\geq \kappa^\lambda \geq 2^\lambda > \kappa$$.

$$2\implies 3$$: Assume $$\kappa$$ is regular and for all $$\lambda < \kappa$$, $$2^\lambda \leq \kappa$$. If $$\kappa$$ is a limit, then it is a regular limit, so it is $$\aleph_0$$ or inaccessible. If $$\kappa = \mu^+$$ is a successor, then $$2^\mu \leq \kappa$$, but also $$2^\mu>\mu$$ implies $$2^\mu\geq \kappa$$. So $$2^\mu = \kappa$$.

$$3\implies 1$$: (a) $$\aleph_0^n = \aleph_0$$ for all finite $$n$$, so $$\aleph_0^{<\aleph_0} = \aleph_0$$.

(b) Suppose $$\kappa$$ is an uncountable regular limit cardinal with $$2^\lambda\leq \kappa$$ for all $$\lambda < \kappa$$. Fix $$\mu$$ with $$\aleph_0\leq \mu<\kappa$$. Since $$\kappa$$ is regular, every function $$\mu\to \kappa$$ has bounded range. So $$\kappa^\mu = \sup_{\mu\leq \lambda < \kappa} \lambda^\mu$$. Now $$\lambda^\mu \leq (2^\lambda)^\lambda = 2^\lambda \leq \kappa$$, so $$\kappa^\mu \leq \sup_{\mu\leq \lambda < \kappa} \kappa = \kappa$$. Thus $$\kappa^{<\kappa} = \kappa$$.

(c) If $$\kappa = \mu^+$$ and $$2^\mu = \kappa$$, then $$\kappa^{<\kappa} = \kappa^\mu = (2^\mu)^\mu = 2^\mu = \kappa$$.

So what the condition $$\kappa^{<\kappa} = \kappa$$ tells us about the "size" of $$\kappa$$ is entirely dependent on the whims of the continuum function.

At one extreme, if GCH holds, then every regular cardinal satisfies $$\kappa^{<\kappa} = \kappa$$.

At the other extreme, it is possible that there are no uncountable cardinals satisfying $$\kappa^{<\kappa} = \kappa$$. It is a theorem of Woodin (see Foreman and Woodin The GCH can fail everywhere) that assuming a certain large cardinal hypothesis is consistent, we can find a model of set theory in which $$2^\kappa = \kappa^{++}$$ for all infinite cardinals $$\kappa$$. In this model, every limit cardinal is a strong limit, so every inaccessible cardinal is strongly inaccessible. Then cutting off the model at $$V_\kappa$$, where $$\kappa$$ is the least inaccessible cardinal in the model, gives a new model in which GCH fails everywhere and there are no inaccessible cardinals, so there are no cardinals satisfying (b) or (c) in the characterization above.

• You need strict inequality to $2^\lambda <\kappa$ to conclude it is a strong limit. It’s possible for such a limit $\kappa$ to be only weakly inaccessible, e.g. start with a model of GCH and blow the continuum up to some inaccessible. Oct 30, 2022 at 18:52
• @spaceisdarkgreen Oh yes, good point. I'll edit. It's too bad - that makes the characterization less snappy. Oct 30, 2022 at 18:56