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 $\lambda \ll\mu$.

| cite | improve this answer | |

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.