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1

No, due to a simple reason: Let $C$ be a club class and let $\mu$ be any infinite cardinal (e.g. $\mu = \omega$). Then there is some $\kappa \in C$ such that $\operatorname{cf}(\kappa) = \operatorname{cf}(\mu)$. Proof. Recursively construct a sequence $(\kappa_{i} \mid i \le \mu)$ be fixing any $\kappa_{0} \in C$, further letting $\kappa_{\alpha+1} := \min ...


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The obvious answer is that you get the consistency of $\sf ZFC$ with an inaccessible. But you get more. You get the existence of a transitive model of this theory. But you actually get more. Since an inaccessible cardinal is the limit of worldly cardinals,1 you get a transitive model of $\sf ZFC$+There exists an inaccessible cardinals+There is a proper class ...


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Let $C(k)$ be the set of club subsets of $k.$ Let $R(k)=\{l\in k: l=cf(l)\}.$ Observe that for any set $S\subset On,$ if $S\in L$ then (1) $\forall a\in On \;[\;\{b\cap S :b\in a\} \in L\;], \; \text {and}$ (2) $\forall a\in On\;[\;a=\cup (a\cap S)\iff L\Vdash (a= \cup (a\cap S)\;].$ (3) Also observe that $\forall a\in On\;[a=|a|\implies L\Vdash a=|a|....


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A "variation" on user203787's proof that $k$ is a strong limit: Let $0<l<k$ and let $F:k\to 2^l.$ For each $x\in l,$ and for $j\in \{0,1\},\; $ let $x_j=\{y\in k\;:F(y)(x)=j\}. $ And define $G\in 2^l$ by $G(x)=j\iff m(x_j)=1.\; $ That is, $ m(x_{G(x)})=1. $ And $\forall y\in x_{G(x)}\;(F(y)(x)=G(x)).$ For brevity let $ \;S=\cap_{x\in l}\; x_{G(x)}.\...


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My personal favorite - although it seems not to be very popular - is the inner model hypothesis (IMH) and its variants. IMH states that anything that can happen in an inner model, does: specifically, $$\mbox{Any parameter-free $\varphi$ which holds in an inner model of some outer model of $V$,}$$ $$\mbox{already holds in some inner model of $V$.}$$ (Of ...



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