Questions tagged [large-cardinals]

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11 votes
1 answer
437 views

Does $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$?

Every $Π^1_1$ formula $φ$ without free second order variables can be converted into a $Σ^1_1$ $ψ$ such that $φ ⇔ ψ^\mathrm{HYP}$, and vice versa. ($\mathrm{HYP}$ is the hyperarithmetical universe, ...
Dmytro Taranovsky's user avatar
1 vote
0 answers
311 views

Is there a second-order expression of '$\kappa$ is Reinhardt' in $NGB$, where $\kappa$ is a cardinal?

In their paper, "Generalizations of the Kunen inconsistency", (Annals of Pure and Applied Logic 163 (2012) 1872-1890, doi:10.1016/j.apal.2012.06.001, arXiv:1106.1951), Hamkins, Kirmayer, and ...
Thomas Benjamin's user avatar
6 votes
0 answers
382 views

General theory of the reals in Solovay-like models

Solovay's model is a famous model of $\sf ZF$ where we start in $L$ with $\kappa$ inaccessible, and we collapse all the ordinals below $\kappa$ to be countable, without collapsing $\kappa$ itself, and ...
Asaf Karagila's user avatar
  • 38.1k
1 vote
0 answers
160 views

A parsimonious large cardinal axiom

The ordering of large cardinals by consistency strength is well known. I was wondering what one can say regarding an ordering by direct implication. In particular, I am looking for is a parsimonious ...
Anindya's user avatar
  • 665
7 votes
0 answers
307 views

$0^\#$ in weak theories vs large cardinals in $L$

To better understand the transition from large cardinal axioms consistent with the constructible universe $L$ to large cardinal axioms transcending $L$, I am looking for natural equiconsistencies ...
Dmytro Taranovsky's user avatar
4 votes
0 answers
190 views

Elementary self-embeddings conservative over ZFC

Question: Is the following theory conservative over ZFC? And if not, what is its strength? Language: $∈$, $j$ (unary function symbol) Axioms: 1. ZFC (without separation and replacement for formulas ...
Dmytro Taranovsky's user avatar
3 votes
1 answer
234 views

The intersection of all normal ultrafilters on a measurable cardinal

Suppose $\kappa$ is a measurable cardinal. Let $W$ be the intersection of all normal ultrafilters on $\kappa$. I am interested in a precise characterization of the filter $W$. One sure way to ...
D. Hershko's user avatar
12 votes
3 answers
1k views

Necessary use of large cardinals in mathematics [duplicate]

There are some statements, whose consistency (or the consistency of their negation) require the existence of large cardinals (in the sense that if the statement (or its negation) is consistent, then ...
14 votes
0 answers
516 views

Seeing what gets Harvey Friedman's "tangible incompleteness" principles into large cardinal territory

I'm trying to wrap my head around some of Harvey Friedman's recent, unpublished work on his tangible incompleteness project, and I'm trying to see the link between his "tangible statements" (...
Malice Vidrine's user avatar
1 vote
1 answer
125 views

Complexity of a proper class of extendibles

If consistent, is existence of a proper class of extendible cardinals provably equivalent to a $Σ^V_5$ statement? Recall that in ZFC, a cardinal $κ$ is extendible iff for every $λ>κ$ there is an ...
Dmytro Taranovsky's user avatar
-1 votes
1 answer
366 views

What is the consistency strength of adding this ordinal reflection scheme on top of Ackermann set theory?

Axiom scheme of Ordinal Reflection: if $\phi$ is a formula that doesn't use the symbol $V$, whose parameters are among $x_1,..,x_n$; then: $$\forall x_1 \in V,\dotsc,\forall x_n \in V: \phi(On) \to \\\...
Zuhair Al-Johar's user avatar
6 votes
1 answer
414 views

When does "sufficient genericity" actually suffice?

Fix a forcing notion $\mathbb{P}$. Say that a formula $\varphi(x)$ with parameters is $\mathbb{P}$-enforceable if there is some countable set $\mathcal{D}$ of dense sets in $\mathbb{P}$ such that for ...
Noah Schweber's user avatar
6 votes
1 answer
811 views

Very large axiom of choice

let me say that I am not a set theorist, but I have to settle up some things in category theory and I need your help. What I'd like to do is, in some way, use axiom of choice for proper classes. I ...
Andrea Marino's user avatar
5 votes
1 answer
539 views

Can the category of partial orders be fully embedded in the category of linear orders?

Q(1): Can the category of partial orders be fully embedded in the category of linear orders? Vopěnka's principle, or VP, is a very intriguing axiom with many equivalent forms and consequences ...
Keith Millar's user avatar
  • 1,242
3 votes
1 answer
443 views

(ZC + $\Sigma_2$ replacement + inaccessible cardinal) equiconsistent with (ZFC + inaccessible cardinal)?

Randall Holmes has made a quite convincing argument against the fact that the full axiom schema of replacement should be considered as “intuitively obvious”—even though he does believe ZFC to be ...
Rémi Peyre's user avatar
0 votes
0 answers
116 views

What's the consistency strength of this theory of Stretchable Hierarchies?

Working in Morse-Kelley set theory: A hierarchy is defined as a class that is the union of sets uniquely indexed by ordinals, called as stages, such that each stage is the power set of the ...
Zuhair Al-Johar's user avatar
-2 votes
1 answer
126 views

Is the principle of indifference of hierarchical construction consistent? What's its consistency strength?

Sometimes when one tries to capture the abstract aspect of some notion that is intuitively considered as being truly of abstract nature, in set theoretic terms, then this can extend the theory in a ...
Zuhair Al-Johar's user avatar
2 votes
1 answer
229 views

Possible inconsistency related to embeddings $j: M\prec V$

In the paper Vickers, J.; Welch, P. D., On elementary embeddings from an inner model to the universe, J. Symb. Log. 66, No. 3, 1090-1116 (2001). ZBL1025.03049. it is stated to that if $Ord$ is ...
Master's user avatar
  • 1,103
0 votes
2 answers
553 views

Why do ordinal collapsing functions use regular cardinals?

Inaccessible cardinals are defined as regular strong limit cardinal, and weakly inaccessible cardinals as regular weak limit cardinal. These cardinals are used by some ordinal collapsing functions. My ...
Jacques's user avatar
0 votes
1 answer
281 views

Is there no anti-foundational theory exists Reinhardt and hold Global Choice?

J. D. Hamkins proved in "The foundation axiom and elementary self-embeddings of the universe" that, working in $ZFGC^− +BAFA$, there are nontrivial automorphisms and elementary embeddings of the ...
Ember Edison's user avatar
0 votes
0 answers
174 views

Can second order ordinal arithmetic be extended to the same extent as ZFC?

In a prior posting, I've posed the idea of reducing set theory to an extended kind of second order ordinal arithmetic $\small \sf`` 2 oO A"$. The idea was to have a domain of ordinals and sets of ...
Zuhair Al-Johar's user avatar
15 votes
1 answer
536 views

What would cohomological localization be good for?

An open problem in algebraic topology is whether arbitrary cohomological localizations of simplicial sets (or, equivalently, topological spaces) can be proven to exist in ZFC. It's provable in ZFC ...
Mike Shulman's user avatar
5 votes
2 answers
594 views

A weak (?) form of Shelah cardinals

The following definition of a large cardinal property combines parts of the definitions of "Shelah cardinal" and "Woodin cardinal": A cardinal $\kappa$ is weakly Shelah if for all $f : \kappa \to \...
Trevor Wilson's user avatar
1 vote
1 answer
334 views

Proving independence with large cardinals?

Suppose I want to prove some statement S is independent of ZFC. Now instead of the usual approach of making models, I do the following: - Take two large cardinal axioms L1 and L2 - Prove that ZFC + L1 ...
Anindya's user avatar
  • 665
8 votes
4 answers
1k views

"Bootstrapping" an unbounded class of inaccessible cardinals

The "richness principle" of set theory asserts roughly that "everything that happens once should happen an unbounded number of times". An example would be the existence of an unbounded class of ...
Anindya's user avatar
  • 665
5 votes
2 answers
288 views

Stationary sets and $\kappa$-complete normal ultrafilters

Let $\kappa$ be a measurable cardinal, and let $u$ be a normal $\kappa$-complete ultrafilter over $\kappa$. It is a standard easy fact that every closed unbounded set must belong to $u$ (notice that ...
David Fernandez-Breton's user avatar
5 votes
1 answer
239 views

Uniqueness of countable version of $L[U]$?

Suppose $\alpha$ is a countable ordinal and $U_0,U_1,\kappa$ are such that $L_\alpha[U_i] \models \mathrm{ZFC} + U_i$ is a normal ultrafilter on $\kappa$. Does $U_0 = U_1$? The argument for ...
Monroe Eskew's user avatar
  • 18.1k
-1 votes
1 answer
131 views

What is the strength of claiming that the class of all $V_\kappa$ stages that are $H_\kappa$ when $\kappa$ is regular, is inaccessible?

[EDIT] This posting had been edited to assert that we are speaking about regular mutual stages. Let $H_{\kappa}$ be the set of all sets that are hereditarily strictly smaller in cardinality than ...
Zuhair Al-Johar's user avatar
4 votes
1 answer
537 views

Upward reflection of rank-into-rank cardinals

Rank-into-rank cardinals have the rather intriguing property that they reflect upwards. I would be interested to know how far the upward reflection goes: 1) Does "There exists a rank-into-rank ...
Anindya's user avatar
  • 665
6 votes
0 answers
173 views

Generalized graph-minor theorem?

Consider the following generalized graph-minor theorem: GM($κ,λ$): Given any collection $S$ of $κ$ simple undirected graphs each with less than $λ$ vertices, there are distinct graphs $G,H$ in $S$ ...
user21820's user avatar
  • 2,734
0 votes
1 answer
409 views

Proving that being an inaccessible cardinal is absolute, for $V_\kappa$, where $\kappa$ is inaccessible?

I'm going through the proof that if $\kappa$ is inaccessible then $V_\kappa \vDash \mathrm{ZFC}$ and how thus we have $\mathrm{ZF} \nvdash \text{"There exist inaccessible cardinals"}$. So the last ...
lost_set_theory_student's user avatar
6 votes
1 answer
232 views

Generic saturation of inner models

Say that an inner model $M$ of $V$ is generically saturated if for every forcing notion $\Bbb P\in M$, either there is an $M$-generic for $\Bbb P$ in $V$, or forcing with $\Bbb P$ over $V$ collapses ...
Asaf Karagila's user avatar
  • 38.1k
12 votes
1 answer
461 views

Does the statement 'there exists a first-order theory $T$ with no saturated models' have any set theoretic strength?

Exercises 6, 7, and 8 in section 10.4 of Hodges' big model theory textbook contain an outline of a proof of the consistency of the following statement There exists a countable first-order theory $T$...
James Hanson's user avatar
  • 10.3k
13 votes
1 answer
737 views

Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncountable cofinality?

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 ...
Asaf Karagila's user avatar
  • 38.1k
13 votes
1 answer
561 views

End-extending cardinals

Let us say a cardinal $\kappa$ end-extending if there is a function $F : V_\kappa^{<\omega} \to V_\kappa$ such that: (a) If $M \subseteq V_\kappa$ is closed under $F$, then $M \prec V_\kappa$. (b) ...
Monroe Eskew's user avatar
  • 18.1k
3 votes
1 answer
233 views

Radin forcing preserving large cardinals

I'm wondering if there are any known result for the maximum large cardinal strength which can be preserved by Radin forcing? For instance, with any large cardinal hypothesis in the ground model, can ...
Jiachen Yuan's user avatar
1 vote
0 answers
90 views

Why does $p_{n}(i,1)=1$ so often where the polynomials $p_{n}$ are obtained from the classical Laver tables

So I was doing some computer calculations with the classical Laver tables and I found polynomials $p_{n}(x,y)$ such that $p_{n}(i,1)=1$ for many $n$. The $n$-th classical Laver table is the unique ...
Joseph Van Name's user avatar
1 vote
0 answers
60 views

Growth rate of the critical points of the Fibonacci terms $t_{n}(x,y)$ vs $t_{n}(1,1)$ in the classical Laver tables

The classical Laver table $A_{n}$ is the unique algebra $(\{1,\dots,2^{n}\},*_{n})$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ and $x*_{n}1=x+1\mod 2^{n}$ for all $x,y,z\in A_{n}$. Define the ...
Joseph Van Name's user avatar
2 votes
1 answer
105 views

Attraction in Laver tables

If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x*x^{[n]}$. The motivation for this question comes from the following fact about ...
Joseph Van Name's user avatar
5 votes
1 answer
241 views

Amalgamation via elementary embeddings

Can there exist three transitive models of ZFC with the same ordinals, $M_0,M_1,N$, such that there are elementary embeddings $j_i : M_i \to N$ for $i<2$, but there is no elementary embedding from $...
Monroe Eskew's user avatar
  • 18.1k
1 vote
0 answers
75 views

Multiple roots in the classical Laver tables

The classical Laver table $A_{n}$ is the unique algebraic structure $$(\{1,\dots,2^{n}\},*_{n})$$ such that $x*_{n}1=x+1\mod 2^{n}$ and $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ for all $x,y,z\in\{1,...
Joseph Van Name's user avatar
1 vote
0 answers
59 views

Can we have $\sup\{\alpha\mid(x*x)^{\sharp}(\alpha)>x^{\sharp}(\alpha)\}=\infty$ in an algebra resembling the algebras of elementary embeddings?

A finite algebra $(X,*,1)$ is a reduced Laver-like algebra if it satisfies the identities $x*(y*z)=(x*y)*(x*z)$ and if there is a surjective function $\mathrm{crit}:X\rightarrow n+1$ where $\mathrm{...
Joseph Van Name's user avatar
1 vote
0 answers
60 views

In the classical Laver tables, do we have $o_{n}(1)<o_{n}(2)$ for any $n>8$?

The classical Laver table $A_{n}$ is the unique algebraic structure $(\{1,\dots,2^{n}\},*_{n})$ where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and where $$x*_{n}1=x+1\mod 2^{n}$$ for $x,y,z\in\{1,\...
Joseph Van Name's user avatar
2 votes
0 answers
81 views

For each $n$ is it possible to have $\mathrm{crit}(x^{[n]}*y)>\mathrm{crit}(x^{[n-1]}*y)>\dots>\mathrm{crit}(x*y)$?

Suppose that $(X,*,1)$ satisfies the following identities: $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$. Define the Fibonacci terms $t_{n}(x,y)$ for $n\geq 1$ by letting $$t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=...
Joseph Van Name's user avatar
1 vote
0 answers
44 views

Vastness of inverse systems of Laver-like algebras

Suppose that $(X,*,1)$ satisfies the identities $x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$. Then we say that $(X,*,1)$ is a reduced Laver-like algebra if whenever $x_{n}\in X$ for all $n\in\omega$, there is ...
Joseph Van Name's user avatar
1 vote
0 answers
33 views

Can we always extend a finitely generated reduced Laver-like algebra to a vast inverse system of Laver-like algebras?

An $(X,*,1)$ that satisfies the identities $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$ is said to be a reduced Laver-like algebra if whenever $x_{n}\in X$ for $n\in\omega$, there is some $N\in\omega$ where $x_{...
Joseph Van Name's user avatar
11 votes
3 answers
1k views

Higher $\infty$-categories

Is there a reason we consider $\infty$-categories to be the $\omega^{th}$ step in the 2-internalization inside Cat (or enrichment over Cat if you prefer)* process made invertible above some finite ...
Alec Rhea's user avatar
  • 8,977
7 votes
1 answer
370 views

Locally presentable categories, universes, and Vopenka's principle

Some aspects of the theory of locally presentable categories depends on set-theoretic assumptions. Namely, a large cardinal axiom known as Vopenka's principle implies several nice properties of such ...
Valery Isaev's user avatar
  • 4,410
3 votes
0 answers
240 views

Ordering large cardinal axioms around the level of $n$-huge by consistency strength?

So the large cardinal axioms are for the most part considered to be linearly ordered by consistency strength. For the large cardinals between extendibility and rank-into-rank (i.e. the $n$-huge ...
Joseph Van Name's user avatar
2 votes
0 answers
53 views

Calibrating the strength of the quotients of subalgebras of the classical Laver tables

Define an algebraic structure $A_{n}$ by letting $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ where $*_{n}$ is the unique operation such that $x*_{n}1=x+1\mod 2^{n}$ for $$x\in\{1,\dots,2^{n}-1,2^{n}\}$...
Joseph Van Name's user avatar

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