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Similar to what Mohammad writes, but slightly different, Magidor found the identity crisis of strongly compact cardinals: It is consistent that the least strongly compact cardinal is the least mea …

Actually, if you assume that every set is inside a universe, then you can get something slightly weaker, but you do get a class of inaccessible cardinals which are regular. So this is already somethin …

Yes. If $\kappa$ is a Worldly cardinal, then $V_\kappa$ is a model of $\sf ZFC$, indeed a transitive model. And therefore $\rm Con\sf (ZFC)$ holds. As you already know, the least Worldly cardinal is …

This is a question from Jech's Set Theory (Ex. 17.12) which I'm reading at the moment and pretty much stuck on. If $D$ is a normal measure on $\kappa$ and $\{ \aleph_\alpha \colon > 2^{\aleph_\a …

Suppose $j\colon V\to M$ is an elementary embedding and $\kappa$ is the critical point of $j$, then $\kappa$ is measurable, and we can define the ultrafilter $U$ over $\kappa$ as: $$A\in U\iff \kappa\ …

No. Work in $L[U]$, the canonical inner model, then $U$ is the unique normal measure on $\kappa$. Pick any $S$ such that $S$ and $\kappa\setminus S$ are stationary, and then only one of them can be in …

It is a theorem of A. Levy, if $\kappa$ is an inaccessible cardinal, then $V_\kappa\prec_{\Sigma_1} V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ sentences. One migh …

It is known that $L\models 2^\kappa=\kappa^+$, and that for a set of ordinals $A$ we know that $L[A]\models \exists\lambda\forall\kappa>\lambda(2^\kappa=\kappa^+)$. In this sense, there is some simil …

Suppose that $\kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^\kappa=\kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent w …

If $\kappa$ is an inaccessible cardinal then $V_\kappa$ is a model of $\sf ZFC_2$ ($\sf ZFC$ with a second-order replacement axiom). If there are many inaccessible cardinals then there are many model …

Consider the axiom of infinity in the axioms of ZF. You cannot prove it holds from the other axioms of ZF; and assuming it means that you can prove the consistency of ZF-Infinity by taking the heredit …

Suppose that $\kappa$ is the critical point of $j\colon V\to M$, and suppose that $\mathcal F=(F_\alpha\mid\alpha\leq\kappa)$ is a sequence such that for every limit ordinal $\alpha$, $F_\alpha$ is a …

Assuming the axiom of choice there is a neat way defining inaccessible cardinals as uncountable, regular, strong limit cardinals. Without the axiom of choice we have several notions of inaccessibilit …

If $\kappa$ is a singular cardinal and we collapse every ordinal below $\kappa$ to be of size $\lambda$, then you might want to say that $\kappa=\lambda^+$. But $\sf ZFC$ proves that $\lambda^+$ has t …

No, yes, and not sure. $H$ (and $H_\kappa$ in general) always satisfies choice, because any family of nonempty sets in $H$ has a well orderable transitive closure, from whence we can define a choice f …