I made a cardinal property as follows
Let $\kappa$ be $\lambda$-Monotonous iff: $$\forall S\in\mathcal{P}_{=\lambda}(\mathrm{V})\forall T\in\mathcal{P}_{=\kappa}(\mathrm{V})(S\subseteq T\rightarrow (S,\in)\prec (T,\in))$$ The intuition behind this is that many $S$ of cardinality $\lambda$ are too similar to tell apart to those of cardinality $T$. So far, I have determined that:
- No Monotonous cardinal $\kappa$ is $\lambda$-Monotonous for any $\lambda\geq\kappa$(This one is fairly obvious)
- No infinite cardinal $\kappa$ is $\lambda$-Monotonous for any finite $\lambda$
- There are no cardinals which are $\aleph_0$-Monotonous, this combined with the last 2 statements means $\aleph_0$ is not Monotonous at all
- Every wordly cardinal is not $\lambda$-Monotonous unless $\lambda$ is wordly
- Every successor to a worldly cardinal is not $\lambda$-Monotonous unless $\lambda$ is a successor to a wordly cardinal
- $\beth_{\omega}$ is not $\lambda$-Monotonous for any $\lambda$ a $\beth$ number (and thus GCH implies $\beth_\omega$ is not Monotonous at all)
Can you find anything else about these cardinals?
-BTW Try using Tarski-Vaught test maybe?
This notion is flat out inconsistent.
Pick $T\prec H(\theta)$ for a large enough $\theta$, then note that $\varnothing\in T$, and pick $S$ to be any subset of $T$ of the right cardinality, then $S\setminus\{\varnothing\}$ has the same cardinality and it is still a subset of $T$. But it is impossible that $S\setminus\{\varnothing\}\prec T$, because $\varnothing$ is definable.