# Raising the index of accessibility

In the standard reference books Locally presentable and accessible categories (Adamek-Rosicky, Theorem 2.11) and Accessible categories (Makkai-Pare, $$\S$$2.3), it is shown that for regular cardinals $$\lambda\le\mu$$, the following are equivalent:

• Every $$\lambda$$-accessible category is $$\mu$$-accessible.
• For every $$\mu'<\mu$$, the set $$P_\lambda(\mu')$$ of subsets of $$\mu'$$ of cardinality $$<\lambda$$ has a cofinal subset of cardinality $$<\mu$$.

This relation is denoted $$\lambda\unlhd \mu$$ (or $$\lambda \lhd \mu$$ for the irreflexive version).

In Higher Topos Theory (Lurie, Definition A.2.6.3) the relation $$\lambda\ll\mu$$ is defined to mean

• For every $$\lambda'<\lambda$$ and $$\mu'<\mu$$ we have $$(\mu')^{\lambda'} <\mu$$.

By Example 2.13(4) of Adamek-Rosicky, if $$\lambda\ll\mu$$ and $$\lambda<\mu$$ then $$\lambda\lhd \mu$$. (Indeed, in this case $$P_\lambda(\mu')$$ itself has cardinality $$<\mu$$.)

Does the converse hold? That is, if $$\lambda\lhd \mu$$ do we have $$\lambda\ll\mu$$?

Note that the two relations definitely differ in the reflexive case: $$\lambda\unlhd\lambda$$ is always true, but $$\lambda\ll\lambda$$ holds only when $$\lambda$$ is inaccessible. It also seems that the converse implies the generalized continuum hypothesis for regular cardinals, since if $$\lambda^+ < 2^\lambda$$ then we have $$\lambda^+ \lhd \lambda^{++}$$ but not $$\lambda^+ \ll \lambda^{++}$$. Thus, the converse is not provable in ZFC. Is it disprovable?

Under GCH, if $$\lambda < \mu$$ are regular cardinals, then $$\lambda \lhd \mu$$ implies $$\lambda \ll\mu$$. The proof uses the following standard fact:

Lemma. Suppose $$\lambda \leq \gamma$$ are infinite cardinals. Then $$\gamma^{<\lambda} = 2^{<\lambda} \cdot \text{cf}(P_\lambda(\gamma))$$.

Proof. Recall that since $$\lambda \leq \gamma$$, $$|P_\lambda(\gamma)| = \gamma^{<\lambda}$$. The inequality $$\gamma^{<\lambda} \geq 2^{<\lambda} \cdot \text{cf}(P_\lambda(\gamma))$$ comes from the fact that $$\gamma^{<\lambda} = |P_\lambda(\gamma)|\geq \text{cf}(P_\lambda(\gamma))$$. Let us prove the reverse inequality. Let $$C\subseteq P_\lambda(\gamma)$$ be a cofinal set of cardinality $$\text{cf}(P_\lambda(\gamma))$$. Note that $$P_\lambda(\gamma) = \bigcup_{\sigma\in C} P(\sigma)$$, so $$\gamma^{<\lambda} = |P_\lambda(\gamma)| \leq |C|\cdot \sup_{\sigma\in C}|P(\sigma)| \leq \text{cf}(P_\lambda(\gamma)) \cdot 2^{<\lambda}$$, as desired.

Under GCH, this lemma has the following corollary:

Corollary (GCH). Suppose $$\lambda \leq \gamma$$ are infinite cardinals and $$\lambda$$ is regular. Then $$\gamma^{<\lambda} = \text{cf}(P_\lambda(\gamma))$$.

Proof. By the lemma, it suffices to show that $$\text{cf}(P_\lambda(\gamma))\geq 2^{<\lambda}$$. By GCH, $$2^{<\lambda} =\lambda$$. But $$\text{cf}(P_\lambda(\gamma)) \geq \text{cf}(P_\lambda(\lambda)) \geq \lambda$$ since $$\lambda$$ is regular.

Proposition (GCH). If $$\lambda < \mu$$ are regular cardinals, then $$\lambda \lhd \mu$$ implies $$\lambda \ll\mu$$

Proof. Suppose $$\lambda < \mu$$ are regular cardinals such that $$\lambda \lhd \mu$$. In other words $$\text{cf}(P_\lambda(\mu')) < \mu$$ for all $$\mu' < \lambda$$. By the corollary, $$(\mu')^{<\lambda} < \mu$$ for all $$\mu' < \mu$$. In particular, $$(\mu')^{\lambda'} < \mu$$ for all $$\mu' < \mu$$ and $$\lambda' < \lambda$$. Thus $$\lambda \ll\mu$$.

More generally, the following is Fact 2.5 of Lieberman, Rosický, and Vasey - Internal sizes in $$\mu$$-abstract elementary classes:

Theorem. Assume $$\lambda$$ and $$\mu$$ are regular cardinals and $$2^{<\lambda} < \mu$$. Then $$\lambda \triangleleft \mu$$ if and only if $$\lambda \ll \mu$$.

Assuming GCH, $$2^{<\lambda} = \lambda$$, so we recover Gabe's answer (the proof is the same).