6
$\begingroup$

Makkai and Paré introduced the following binary relation on regular cardinals: given $\kappa$ and $\lambda$, $\kappa \vartriangleleft \lambda$ (read, $\kappa$ is sharply less than $\lambda$) when $\kappa < \lambda$ and, for every set $X$ of cardinality $< \lambda$, the set $P_\kappa (X)$ of all subsets of $X$ of cardinality $< \kappa$ has a cofinal subset of cardinality $< \lambda$.

It is not hard to see that $\kappa \vartriangleleft \kappa^+$ for all regular cardinals $\kappa$. On the other hand, $\aleph_1$ is not sharply less than $\aleph_{\omega + 1}$, so $\vartriangleleft$ is not the same as $<$. Nonetheless, it is true that $\aleph_0 \vartriangleleft \lambda$ for every uncountable regular cardinal $\lambda$, simply because $P_{\aleph_0} (X)$ has the same cardinality as $X$ when $X$ is infinite. More generally, if for all (not necessarily regular) cardinals $\kappa' < \kappa$ and all cardinals $\lambda' < \lambda$, we have ${\lambda'}^{\kappa'} < \lambda$, then $\kappa \vartriangleleft \lambda$. In particular if $\lambda$ is an inaccessible cardinal then $\kappa \vartriangleleft \lambda$ for all regular cardinals $\kappa < \lambda$.

Question. Do there exist uncountable regular cardinals $\kappa$ such that $\kappa \vartriangleleft \lambda$ if and only if $\kappa < \lambda$? Is there a proper class of them?

$\endgroup$
0

2 Answers 2

4
$\begingroup$

First note that if $\kappa \vartriangleleft \mu^+$ then $\mu^{\lt\kappa} = |P_\kappa(\mu)| \leq 2^{\lt\kappa}\cdot\mu$. Therefore $\mu^{\lt\kappa} = \mu$ if $\kappa \vartriangleleft \mu^+$ and $2^{\lt\kappa} \leq \mu$. This is impossible if $\operatorname{cf}(\mu) \lt \kappa$ by König's Theorem, which implies that $\mu^{\operatorname{cf}(\mu)} \gt \mu$ always holds. Since there are arbitrarily large cardinals $\mu$ with countable cofinality, the only infinite cardinal with this property is $\aleph_0$.

$\endgroup$
2
  • $\begingroup$ I'm confused. The relation $\vartriangleleft$ is only defined for regular cardinals, but an uncountable cardinal $\lambda$ with countable cofinality is necessarily singular and therefore exempt from the question, no? $\endgroup$
    – Zhen Lin
    Nov 29, 2013 at 15:53
  • $\begingroup$ @ZhenLin: Actually, I meant $\kappa \vartriangleleft \lambda^+$ in the original. I'm now using $\mu$ to avoid confusion. $\endgroup$ Nov 29, 2013 at 17:07
4
$\begingroup$

No.

Given regular cardinals $\kappa$ and $\lambda$, $\kappa \vartriangleleft \lambda$ iff $\kappa < \lambda$ and $$ \operatorname{cf} ([ \mu ]^{< \kappa} , \subseteq) = \operatorname{cov} (\mu , \kappa , \kappa) < \lambda $$ for every cardinal $\mu$ with $\kappa \leq \mu < \lambda$.

Let $\kappa$ be any regular cardinal > $\aleph_0$. Then, let $\delta$ be the ordinal such that $\kappa = \aleph_{\delta}$. Define $\lambda = \aleph_{\delta + \omega + 1}$.

Now, $$ \operatorname{cov} (\aleph_{\delta + \omega} , \kappa , \kappa) \geq $$ $$ \operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , \aleph_1) = $$ $$ \operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , {(\operatorname{cf} (\aleph_{\delta + \omega}))}^+) \geq $$ $$ \operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , {(\operatorname{cf} (\aleph_{\delta + \omega}))}^+ , \operatorname{cf} (\aleph_{\delta + \omega})) \geq $$ $$ \aleph_{\delta + \omega + 1} = \lambda , $$ the last inequality by Fact 1 in

Andreas Liu, Bounds for covering numbers, The Journal of Symbolic Logic 71 (2006), 1303-1310. doi:10.2178/jsl/1164060456

Hence, $\kappa < \lambda$ but $\neg (\kappa \vartriangleleft \lambda)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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