If $\lambda, \mu$ are regular cardinals, say that $\lambda \trianglelefteq \mu$ if $\lambda \leq \mu$ and $$\forall X, \, |X| < \mu \implies \mathrm{cf} (P_\lambda(X)) < \mu$$ Here $P_\lambda(X)$ is the poset of subsets of $X$ of cardinality $<\lambda$, and $\mathrm{cf}$ takes the cofinality of a poset.

Then $\trianglelefteq$ is an $\mathrm{Ord}$-directed partial order on regular cardinals which is strictly weaker than the ordering by size (by $\mathrm{Ord}$-directed, I mean that any set of cardinals has a $\triangleleft$-upper bound). For instance, $\aleph_1 \not \trianglelefteq \aleph_{\omega+1}$ but $\aleph_0 \trianglelefteq \mu$ for all $\mu \geq \aleph_0$.

This relation comes up in the theory of accessible categories, because the following are equivalent when $\lambda \leq \mu$ are regular cardinals:

  • Every $\lambda$-accessible category is $\mu$-accessible.
  • The category $\mathrm{Pos}_\lambda$ of $\lambda$-directed posets and embeddings [which is always $\lambda$-accessible] is $\mu$-accessible.
  • For every $\lambda$-directed poset $P$, the $\lambda$-directed subsets are cofinal in $P_\mu(P)$.
  • $\lambda \trianglelefteq \mu$.

A good reference is Adámek and Rosický, though these facts mostly go back at least to Makkai and Paré. The only facts I know how to use to determine whether $\lambda \triangleleft \mu$ are the following:

  1. $\lambda \triangleleft \lambda^+$
  2. If $\left(\alpha <\lambda, \beta < \mu \implies \beta^\alpha < \mu\right)$, then $\lambda \triangleleft \mu$ for the somewhat boring reason that for $|X| < \mu$ we have $|P_\lambda(X)| < \mu$ so that certainly $\mathrm{cf} (P_\lambda(X)) < \mu$.
  3. If $\lambda < \mu$ and $\mathrm{cf} \mu \leq \lambda$, then $\lambda^+ \not \triangleleft \mu^+$.

Questions I basically would just like to see what the set theorists (and category theorists) around here might have to say about the relation $\triangleleft$. Is this by chance a well-studied relation? Should I expect the question of whether $\lambda \triangleleft \mu$ holds to be mostly determined, or mostly independent, given reasonable descriptions of $\lambda$ and $\mu$? Should I expect (1), (2), and (3) to pretty much determine what can be determined, or should I expect there's more to say? I suppose things are not too complicated under GCH, for example. Does having good control over the $\beth$ function in general give good control over the functions $\mathrm{cf}(P_\lambda(\beta))$?

A really basic question: is the quantity $\mathrm{cf}(P_\lambda(X))$ (weakly) monotonic in $|X|$?

  • 1
    $\begingroup$ I added the [lo.logic] tag, since people who might be interested in this question might be watching that tag. $\endgroup$ – David Roberts Jul 1 '16 at 1:11
  • 4
    $\begingroup$ Gosh, this gets really deep really fast! The basic stuff is well known, especially assuming GCH, but PCF Theory is based on the idea that one can calculate bounds on complicated stuff like $\operatorname{cf}(P_{\aleph_0}(\aleph_\omega))$ in ZFC. I hope somebody who is on top of things comes around and gives a thoughtful answer... $\endgroup$ – François G. Dorais Jul 1 '16 at 5:54
  • 1
    $\begingroup$ @FrançoisG.Dorais I was afraid to ask whether PCF theory might be relevant, since on the face of it I'm talking about the cofinalities of rather different posets. I'd be interested to learn about the nature of this connection. $\endgroup$ – Tim Campion Jul 1 '16 at 15:07
  • $\begingroup$ Some interesting information about the this relation is to be found in the answers to this question. $\endgroup$ – Tim Campion Apr 3 '19 at 1:11

For your really basic question at the end: Yes, the cofinality of $P_\lambda(X)$ is weakly monotone in $X$.

Specifically, if $X\subseteq Y$, then $\newcommand\cof{\text{cof}}\cof(P_\lambda(X))\leq\cof(P_\lambda(Y))$. The reason is that if $A\subset P_\lambda(Y)$ is cofinal in $P_\lambda(Y)$, then the set $A\upharpoonright X=\{a\cap X\mid a\in A\}$ is cofinal in $P_\lambda(X)$.

(Unfortunately, I don't know much about your relation $\trianglelefteq$.)


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.