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Fix a flat pairing function $p : V\times V\to V$. For any infinite ordinal $\alpha$, if $S$ is a binary relation on $V_\alpha$, let $A_S = p[S]$, which is an element of $V_{\alpha+1}$ coding $S$. The …

If $\alpha$ is a limit of $2^\alpha$-supercompact cardinals, then by the Magidor characterization of supercompactness, for each $2^\alpha$-supercompact cardinal $\kappa < \alpha$, for some $\gamma < \ …

Almost $\omega$-huge is equivalent to $\omega$-huge, so it is inconsistent with AC. Closure under $\kappa_n$-sequences plus closure under $\omega$-sequences implies closure under $\delta$-sequences: g …

Suppose $\delta$ is an ordinal. We first note that $\delta$ is an $I_1$-tower cardinal if and only if it has the following superficially weaker property: for all $X\subseteq V_\delta$ there is some $\ …

The answer is no, it is not consistent. If $j : V_{\lambda+2}\to V_{\lambda+2}$ is $\Sigma_1$-elementary, then $\mathcal U = \{A\subseteq P(V_\lambda) : j[V_\lambda]\in j(A)\}$ is a normal fine ultraf …

If $2^\gamma \leq \kappa$ for all $\gamma < \kappa$ (e.g., if GCH holds and $\kappa$ is the least $\aleph_1$-strongly compact), then $|j(\kappa)| \geq 2^\lambda$. To see this, let $\sigma = [\text{id} …

There can be no such set $\mathcal A$. In fact, no set $\mathcal A$ with $j(\mathcal A) = \mathcal A$ and $j(x) = x$ for all $x\in \mathcal A$ can surject onto $\kappa$, and this does not require that …

It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete u …

$2$-extendibility reflects $2^\kappa$-supercompactness. It suffices to show that any $2$-extendible $\kappa$ is $2^\kappa$-supercompact. Then if $\mathcal U$ is a normal fine $\kappa$-complete ultrafi …

No, a Reinhardt cardinal implies SVC is false. First, if there is a Reinhardt cardinal, then by Woodin's proof of the Kunen inconsistency theorem, for sufficiently large regular cardinals $\delta$, th …

To answer the first three questions negatively, the key is to show that measurable weakly Shelah cardinals are limits of weakly Shelah cardinals. To see this, suppose that $\kappa$ is weakly Shelah a …

I prove several new consequences of Reinhardt cardinals in my paper "Measurable cardinals and choiceless axioms," some of which are not known to be consistent relative to any weaker theory. TLDR: Theo …

You will always have $N = M$ and $k = \text{id}$. As Joel mentions, this uses Solovay's lemma that $M = H^M(\text{ran}(j)\cup \{\sup j[\lambda]\})$. We can use this to show that $k$ is surjective, by …

In any model of ZF, there is a surjection from the reals onto $\omega_1$. So if the $\omega_1$ of $L[x]$ (i.e., $(\omega_1)^{L[x]}$) is uncountable, there is a surjection from the reals of $L[x]$ (i.e …

No, if the existence of a Reinhardt is consistent, then it is consistent with a Reinhardt cardinal that the class of inaccessible cardinals is bounded in the ordinals. Indeed, if $j : V\to V$ is a non …