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3
votes
Finite axiomatizability of $\mathrm{WA}_0$
(This should fix an earlier attempt...) I don't see how one can prove that $V$ is the union of that particular chain, but here is a variant, from which we still get finite axiomatizability.
Suppose f …
4
votes
Accepted
Class-theoretic sentences that are $\Pi^1_1$ or $\Pi^1_2$
Much as described by @ElliotGlazer above, let's consider the question for models of the form $(V_\kappa,V_{\kappa+1})$, assuming ZFC in the background, considering all such where $\kappa$ is inaccessi …
4
votes
Accepted
Measurable cardinals from saturated ideals
Yes. And if the $\mathscr{I}_k$s are normal, then the suggested candidate works (I haven't really thought about whether the candidate still works without normality). Here is the argument, which is a t …
5
votes
Stationary vs measurable limits for large cardinals
If there is a normal measure $U$ on $\kappa$ such that $A_X\in U$, where $A_X$ is the set of $X$-cardinals $\alpha<\kappa$, then not only is $\kappa$ a limit of inaccessiles $\kappa'$ such that $\kapp …
6
votes
Accepted
On a question about ordinals $\xi$ satisfying $j_0(\xi)=j_1(\xi)$ for an $I_3$-embedding $j$
Yes. In fact for all ordinals $\xi\notin\mathrm{range}(j_0)$, we have $j_1(\xi)<j_0(\xi)$. For let $\xi$ be an ordinal such that $\xi\notin\mathrm{range}(j_0)$. Let $\gamma$ be least such that $j_0(\g …
10
votes
Can there be an embedding j:V → L, from the set-theoretic universe V to the constructible un...
Theorem 1. If $V$ is a non-trivial set generic extension of $W\models\mathrm{ZFC}$ then there is no
$j:V\to W$ as described (i.e. with $x\in y\iff j(x)\in j(y)$ for all $x,y\in V$).
(So in particular, …
9
votes
Accepted
What is the consistency strength of "Singular worldly that is inaccessible in an inner model"?
To summarize the comments, the hypothesis is equiconsistent with (ZFC +) $0^\sharp$ exists + "there is a worldly cardinal". For as mentioned in the original post, under the hypothesis, $0^\sharp$ exi …
9
votes
Accepted
Weakly compact cardinals in $L$: how long do branches take to appear?
Here is a characterization of $\lambda$, but of course your question is somewhat ambiguous as to what counts as an answer.
For $n<\omega$, let $\alpha_n$ be the least ordinal $\alpha$ such that $\kapp …
5
votes
Accepted
Reinhardt cardinals and iterability
Note that $M=\bigcap_{\alpha}M_\alpha$ is just the $\mathrm{OR}^{\mathrm{th}}$ iterate $M_{\mathrm{OR}}$ cut off at height $\mathrm{OR}$, so we have for example $V_\kappa\preceq V_\lambda\preceq M$ wh …
4
votes
Accepted
Lowenheim-Skolem numbers for SOL + correctness quantifiers
Assume ZFC. Let $n$ be a meta-integer. Then $\kappa$ is $\mathcal{L}^2_n$-correct iff $\kappa$ is $\Sigma_{n+2}$-reflecting, i.e. $V_\kappa\preccurlyeq_{n+2}V$.
Proof: For $n=0$, i.e. 2nd order logic …
9
votes
Accepted
Analogues of worldly cardinals for (an unusual version of) second-order $\mathsf{ZFC}$
Complementing @JasonChen's answer: Assume ZFC+$I_1$ and let $j:V_{\lambda+1}\to V_{\lambda+1}$ be elementary, so $\lambda$ is the sup of the critical sequence of $j$. Then $V_{\lambda}$ models $\mathf …
5
votes
Accepted
Fragility of large cardinals with respect to transitive end extensions
Question 1: Yes, in fact if $M\models$ZFC+"There is a transitive model of $T$" where $T$ is the theory ZFC + "There are two distinct inaccessibles", then there is such an $(A,\alpha)$ - just let $A$ b …
17
votes
A Löwenheim–Skolem–Tarski-like property
Here is an upper bound:
Suppose $\kappa$ is $2$-fold supercompact. Then the property holds at $\kappa$. (Recall that $2$-fold supcompactness means that for each ordinal $\lambda$, there is $j:V\to M$ …
10
votes
Determinacy and Woodin cardinals
(This answers what was the original question, which was with $\mathrm{Coll}(\omega,{<\kappa})$ replacing $\mathrm{Coll}(\omega,\kappa)$.)
Hmm...doesn't this contradict Theorem 1.22 of "MICE WITH FINIT …
5
votes
Accepted
Can this semi-constructible structure satisfy existence of a measurable cardinal?
No. Under the hypotheses, there are limit ordinals $\eta$ such that $\mathcal{P}(\eta)\cap L\subseteq\mathfrak{L}_{\eta+1}$, and therefore, for example, $L_{\eta+2}\cap\mathfrak{L}_{\eta+1}\neq L_{\et …