When are all greater cardinals sharply greater?

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?

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 $\aleph_0$.

• 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? – Zhen Lin Nov 29 '13 at 15:53
• @ZhenLin: Actually, I meant $\kappa \vartriangleleft \lambda^+$ in the original. I'm now using $\mu$ to avoid confusion. – François G. Dorais Nov 29 '13 at 17:07

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 $\neg (\kappa \vartriangleleft \lambda)$.