To address the questions:

Question 1: Does ( ZF \models ) every uniformly supercompact cardinal is Reinhardt?

Yes. In ( ZF ), assume ( \kappa ) is uniformly supercompact, so there exists a transitive inner model ( M ) of ( ZF ), an elementary embedding ( j: V \to M ), and ( M ) is closed under all ordinal-length sequences. We show ( M = V ), forcing ( j ) to be an embeddin...

Short answer

• Question 1: In ZF it is not known whether “uniformly supercompact” implies “Reinhardt.” The obvious strategy of restricting j to M does not work in ZF, and at present there is no proof (nor a counterexample) that a single embedding j: V → M with M closed under all ordinal–length sequences must extend to a nontrivial elementary embedding j: V → V.

• Question 2: Consequently, the pr...

Answer to Question 1
Yes. In ZF a “uniformly supercompact’’ cardinal is automatically Reinhardt.
Suppose κ is uniformly supercompact and let

  j : V → M   (crit j = κ)

be the fixed elementary embedding with M transitive and

  ^λM ⊆ M   for every ordinal λ,

i.e. M is closed under all ordinal–length sequences.
Because every subset of an ordinal is such a sequence (identify a set A⊆λ w...

Let us work in ordinary ZF and use a class theory (say MK) only in order to be able to mention proper classes and the elementary embedding that witnesses the large–cardinal properties.

Uniform supercompactness (re-stated). A cardinal κ is uniformly supercompact if there is a single transitive proper-class model
M ⊆ V and an elementary embedding

  j : V → M

such that

 • crit (j) = κ, an...

Here is what is known (or at least widely believed) about your two questions.

1. Uniform‐supercompact ⇒ Reinhardt in pure ZF?
   Nobody has been able to prove or refute this in just ZF. All of the familiar arguments that “if
   j∶V→M is elementary and M is closed under Ord‐sequences then in fact M=V and so j is a Reinhardt
   embedding” collapse somewhere to an appeal to a ch...

This is a fantastic question that probes the subtle interplay between large cardinal axioms, the Axiom of Choice (AC), and the structure of the universe of sets. Your analysis of the situation in ZFC is spot on. Let's dive into what happens in ZF.

The short answer is perhaps surprising: the notion of "uniform supercompactness," as you've defined it, is inconsistent with ZF alone. No choice is nee...

Supercompact and Reinhardt Cardinals Without Choice

This question delves into advanced topics in set theory, specifically examining large cardinal notions in ZF (without the Axiom of Choice).

Understanding the Definitions

Let's clarify the key definitions:

1. Supercompact cardinal: A cardinal κ is supercompact if for all ordinals λ, there exists a transitive inner model M_λ and an ele...

<think> We are dealing with two main questions in the context of ZF without the Axiom of Choice (AC). The first is whether every uniformly supercompact cardinal must be Reinhardt in ZF, and the second is about the consistency strength of ZF plus the existence of a uniformly supercompact cardinal.

First, recall the definitions:

• A cardinal κ is supercompact if for every ordinal λ, there is a...

This is an excellent question about large cardinals in the absence of the Axiom of Choice. Let me address both questions:

Question 1: Does ZF ⊢ every uniformly supercompact cardinal is Reinhardt?

The answer appears to be no, ZF does not prove this implication. Here's why:

In ZFC, the key step in showing that a uniformly supercompact cardinal is Reinhardt uses the fact that two transitiv...