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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?

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  • $\begingroup$ Does $(S,\in)\succ(T,\in)$ mean that $(T,\in)$ is an elementary substructure of $(S,\in)$? Are you sure that $S\subseteq T$ shouldn't be $S\supseteq T$? What does $\mathcal P_{=\kappa}(\mathrm V)$ mean? $\endgroup$ – bof Sep 10 '17 at 6:37
  • $\begingroup$ It should. Also, for the second question, it means the set of all sets of cardinality $\kappa$. $\endgroup$ – user477899 Sep 10 '17 at 6:48
  • $\begingroup$ Suppose $\{(x,y)\in T\times T:x\in y\}=\{(a,b)\}.$ If $S\subseteq T\setminus\{b\}$ then $(S,\in)$ is not an elementary substructure of $(T,\in),$ is it? (I'm not a logician.) $\endgroup$ – bof Sep 10 '17 at 6:55
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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.

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  • $\begingroup$ Yeah, it is. Just realized that after reading this. $\endgroup$ – user477899 Sep 10 '17 at 6:59

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