For regular cardinals $\lambda\le\mu$, the following are equivalent:
For a proof, see Theorem 2.11 of Adamek-Rosicky or section 2.3 of Makkai-Pare.
If these equivalent conditions hold, we write $\lambda\unlhd \mu$. If $\lambda \unlhd \mu$ and $\lambda\lt\mu$, we write $\lambda\lhd \mu$ and say that $\lambda$ is sharply smaller than $\mu$.
For any uncountable regular cardinal $\lambda$ we have $\aleph_0\lhd \lambda$. (In fact $\aleph_0$ is the only infinite regular cardinal with this property; see this question.)
For any regular cardinal $\lambda$ we have $\lambda\lhd \lambda^+$ (its successor cardinal).
If $\lambda\le\mu$ then $\lambda \lhd (2^\mu)^+$. Thus, for any set $S$ of regular cardinals there is a regular cardinal $\mu$ such that $\lambda\lhd \mu$ for all $\lambda\in S$.
We write $\lambda\ll\mu$ if for every $\lambda'\lt\lambda$ and $\mu'\lt\mu$ we have $(\mu')^{\lambda'} \lt\mu$. (This is Higher Topos Theory, Definition A.2.6.3.) Then if $\lambda\ll\mu$, then $\lambda\lhd \mu$.
The converse claim ($\lambda \lhd \mu$ implies $\lambda\ll\mu$) is independent of ZFC. On one hand it implies the generalized continuum hypothesis (GCH) for regular cardinals (and in particular the ordinary continuum hypothesis (CH)), since if $\lambda^+ \lt 2^\lambda$ then we have $\lambda^+ \lhd \lambda^{++}$ but not $\lambda^+ \ll \lambda^{++}$. Thus it is unprovable in ZFC (if ZFC is consistent), since CH is unprovable. On the other hand, it is implied by the full GCH, as explained by Goldberg, and is thus consistent with ZFC since GCH is.
If $\kappa$ is an inaccessible cardinal, then every $\lambda\lt\kappa$ satisfies $\lambda\lhd \kappa$.
$\aleph_1 \lhd \aleph_{\omega+1}$ does not hold.
Gabe Goldberg, answer to MathOverflow question Raising the index of accessibility, https://mathoverflow.net/q/324492
Last revised on March 14, 2019 at 20:35:01. See the history of this page for a list of all contributions to it.