Questions tagged [large-cardinals]

Large cardinals are such cardinals whose existence cannot be proved within ZFC, and requires stronger axioms to be added to ZFC, they are often used to measure the consistency strength of a certain statement in the language of set theory.

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Sets of Fixed Points in an Ultrapower Embedding

Let $\kappa$ be a measurable cardinal and $j: V \rightarrow M$ be the elementary embedding with critical point $\kappa$. Let $X$ be a set such that for all $x\in X$, $j(x) = x$. My question is: when ...
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$0^\sharp$ and transcendence over $L$

I've started to study the $0^\sharp$ principle from Kanamori's The Higher Infinite, and there are a few interesting, yet a bit vague, remarks scattered across the text that intrigue me and whose core ...
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1answer
135 views

Characterizing worldly cardinals as the supremum (by n) of least $\Sigma_n$-extendible cardinals

A cardinal $\kappa$ is called $\Sigma_n$-extendible iff there is some $\theta$ such that $V_\kappa\prec_{n} V_\theta$. That is, they agree on the truth of $\Sigma_n$ assertions with parameters from $...
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Sizes of consecutive elements of the indiscernible club of $a$, get collapsed to the same cardinality in $L[a^\#]$

This question arises from a proof in Kanamori's "The Higher Infinite". To get started, assume that $a^\#$ exists for every $a \in \omega^\omega$. Also for any $a \in \omega^\omega$, let $I_a$...
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Why ${\rm Ult}(V,{\cal U})\vDash|[id]_{\cal U}|<j_{\cal U}(\kappa)$, when $\cal U$ is a $\delta$-complete fine ultrafilter on $\cal P_\kappa(\alpha)$?

The following argument appears in the proof of Theorem 4.7. in Bagaria-Magidor's paper Group radicals and strongly compact cardinals. Let $\delta<\kappa$ be uncountable cardinals which may be ...
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$\neg \text{CH}$ within an inner model

I know that $\text{ZFC}$ cannot prove the consistency of $\text{ZFC} + \neg \text{CH}$ through inner models, since $\text{V} = \text{L}$ is consistent with $\text{ZFC}$ and within the constructible ...
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Troubles with Kanamori's section on $0^{\sharp}$

I am having trouble understanding the following lemma in Kanamori's book. Some notations: $\mathcal{M}(T,\alpha)$ denotes the (unique up to isomorphism) model of $T$ generated by a set of ...
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2answers
115 views

Is it provable in ZF that there is no nontrivial elementary embedding $\pi: V\to V$?

Is it open whether it can be proved in $ZF$ alone that if $\pi:V\to M$ is a nontrivial elementary embedding, then $M\not=V$?
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Does the Consistency of ZFC imply $V_\kappa$?

It is well known that the existence of an inaccessible cardinal implies the consistency of ZFC. However, I am curious about the converse. Does the consistency of ZFC (plus ZFC) imply the existence of ...
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Why is the status of the period in Laver tables not settled?

Sequence A098820 in the OEIS is purely combinatorially defined (all you need to know is integers), and non-decreasing; it is known (Laver, 95) that assuming Large Cardinal axiom I3 the sequence goes ...
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What large cardinals are limits of Ramsey cardinals?

What large cardinal axiom, if any, implies that there are unboundedly many Ramsey cardinals below the large cardinal? That is, is there a large cardinal axiom, such that if $\kappa$ is that large ...
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What's the order type of the least good pair with that additional qualification?

In this posting I'll add a further condition on good pairs which were defined in a prior posting, I'll re-iterate the definition with the added condition: A good pair $\langle A,B \rangle$ is an ...
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1answer
57 views

What's the limit to formation of ordinals using size-bounded good pairs?

for the same question in a prior posting, add the following size-bounding condition, that is: any subset of $A$ that is equal in size to a proper initial segment of $A$, is bounded in $A$ (with ...
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What is an weakly compact cardinal? [closed]

In set theory there are many types of cardinal numbers. One of them is the weakly compact cardinal. Wikipedia states the following about it: Formally, a cardinal $\kappa$ is defined to be weakly ...
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If $\kappa$ is the $\alpha<\kappa$-th inaccessible cardinal, then the regular cardinals below $\kappa$ form a non-stationary set

How can I prove that if $\kappa$ is the $\alpha<\kappa$-th inaccessible cardinal, then the set of all regular cardinals below $\kappa$ is nonstationary? Is it because the set $\{x: x\text{ is a ...
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If $\kappa$ is ineffable, then so is $\{\nu \lt \kappa: \nu \text{ is weakly-compact}\}$.

So there was an exercise that I solved sometime ago which required me to prove that if $\kappa$ is ineffable, then $\{\nu \lt \kappa: \nu \text{ is weakly-compact}\}$ is stationary. At the time, I ...
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Definability of $j:V\to M$ is an elementary embedding

I was looking at Woodin's chapter in the book Infinity: New Research Frontiers and I'm confused by some of his remarks regarding class-sized elementary embeddings. The following is taken from the book:...
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What can we do with nonenumerable sets of formulas (e.g. formulas of Higher order Logic)?

It is well known textbook fact, that the set of (grammatically correct) sentences/formulas of higher order logic (even of the second order logic) are not enumerable. My question is - what can we do ...
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The least universal cardinal has countable cofinality

We say that a cardinal $\kappa$ is universal if and only if $V_\kappa\models ZFC$. I need to prove that the least universal cardinal has countable cofinality. My attemp: We can list $ZFC:=\{\phi_n:n\...
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1answer
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If $\kappa$ is ineffable, then there is no $\kappa$-Kurepa tree

So this is an exercise($4.25$) from Ralf Schindler's book, and I have some trouble with it. This is the statement: (Jensen-Kunen) Show that if $\kappa$ is ineffable, then there is no $\kappa$-...
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Weak compactness equivalent

I was looking at some exercises in Ralf Schindlers book and I came to an equivalent of weak compactness which to me it seems like it is too weak to be true. So as it stands, the exercise in the book ...
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1answer
62 views

In this model extension ever holds CH

Let $\kappa$ be a measurable cardinal and consider the collapsing forcing $\mathbb{P}:=Coll(\omega_1,<\kappa)$. I know that for every generic extension $V^{\mathbb{P}}$ we get $\kappa=\aleph_2$. ...
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92 views

Inaccessible cardinal in a transitive model

If $\mathfrak M$ is a transitive model of ZFC and $\kappa$ is inaccessible, then $\kappa$ is inaccessible in $\mathfrak M$. Proof. If $\alpha<\kappa$ then since $\mathfrak M \models$ AC, we must ...
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Showing that extenders do not have nice closure properties

This is an exercise from Kanamori's "The Higher Infinite" which I solved by strengthening the hypothesis and I would like to know how we can solve it without doing that. So the exercise is: ...
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the k-SRP stationary Ramsey property

I would like to know why here we have $$\kappa<\omega$$ in "where we partition all of $\kappa<\omega$"
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What are the implications of an infinite series of inaccessible cardinals?

I saw a question where it was asked of the implications of two inaccessible cardinals and wondered about the implications of an infinite series of inaccessible cardinals. I also am wondering of how ...
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1answer
82 views

Equivalence of some statements regarding Vopenka's principle and some $V_\kappa$ for inaccessible $\kappa$

This question of mine arises from Kanamori's The Higher Infinite, where he introduces Vopenka's principle. Let me give some definitions first(assume $\kappa$ is inaccessible throughout this post): A ...
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1answer
101 views

Coding a well-ordering of $2^{|\gamma|^{\lt \kappa}}$ in $P(P(P_\kappa\gamma))$ while preserving relative constructible hierarchy

The title might be a little bit misleading because it is my main question but the setting is in large cardinal theory. But I had no better idea. So to introduce some notation, for any normal ...
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Supercompactness and $1$-extendibility imply we have a club of supercompacts below $\kappa$

My question is about the proof of the following proposition from Kanamori's "The Higher Infinite": Proposition. If $\kappa$ is supercompact and $1$-extendible, then there is a normal ultrafilter ...
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1answer
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The Ulam matrix and the definition of measurable cardinals.

I believe I have a pretty good handle on what an "Ulam matrix" is after reading Ulam's famous 1930 paper on the subject as well as looking at some of the previous answers to questions about that ...
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77 views

Solovay's theorem proof

Question from Jech, Set Theory. In Lemma 20.9 $κ$ is given strongly compact and $λ>κ$ is regular so there exist a fine measure $U$ on $Pκ(λ)$. also, define: $$ [f]_U=lim_{γ<λ}\ \ j_U(γ) $$ and ...
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canonical elementary embedding

Question from Jech, Set Theory- let $ M=Ult(V,U)$ be the Ultrapower of V by U (isomorphic to transitive collapse). And the embedding $$ j:V→M ,j(x):=[c_x] $$ s.t for all $a$ : $[c_x](a) = x$ Lemma ...
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2answers
101 views

Weakly inaccessible cardinal equivalent to regular aleph fixed point?

Are the following propositions equivalent ? $\kappa$ is a weakly inaccessible cardinal $\aleph_\kappa = \kappa \land cof(\kappa) = \kappa$
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Reduction of the satisfaction relation for iterated ultrapowers to that of the first stage and formulas concerning finite product ultrafilters.

There is this lemma in Kanamori's book in the iterated ultrapowers section, which to me seems an important part of the theory, and I don't get part of the proof. So I decided to ask here. Just to ...
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121 views

Does “diagonalization” over elementary embeddings yield another elementary embedding

Suppose we have sequences $\zeta_i^n$ with supremums $\lambda_n$, and elementary embeddings $e_{i,j}^n: V_{\zeta_i^n}\prec V_{\zeta_i^n}$ with critical point $\kappa_i$. The embeddings $e_{i,i+1}^n$ ...
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81 views

Inaccessible cardinal and well founded set

How can I prove the following fact? Let be a regular cardinal $\kappa$ such that $V_\kappa$ is a model of ZFC. Then $\kappa$ is inaccessible.
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Why does an inaccessible cardinal satisfy replacement?

If $\alpha$ is an inaccessible cardinal and $V_{\alpha}$ the corresponding von Neumann universe then $V_{\alpha}$ is supposed to be a model of ZFC. But the singleton $\{V_{\alpha}\}$ is not in $V_{\...
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Ramsey cardinals lemma

I am reading the chapter on large cardinals in Jech's book and I found the following lemma: If $\kappa \rightarrow (\kappa)^{<\omega} $ and if $\lambda < \kappa $ is a cardinal, then $ \kappa \...
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1answer
76 views

How Vopenka's principle implies Semi-Weak Vopenkas principle?

Let's consider 2 large cardinal axioms, the Vopenka's principle (VP) which says that Ord cannot be fully embedded into Graphs and SWVP which says, that the equivalence $Hom(G(\alpha),G(\alpha'))=\...
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1answer
48 views

(S)WVP - Vopenka's priciple

I would like to understand why precisely - in the nice result here on Vopenka's principle - on the 2nd page, in the last but one paragraph it is enough that there is an ordinal $\kappa$ and a ...
4
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1answer
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Isomorphism of $V_\lambda$ and the ultraproduct of $V_{\lambda_{\mathrm{otp}(x)}}, x \in \mathcal{P}_\kappa({\lambda})$ by a normal fine measure.

This question is actually exercise (20.5) from Set Theory by Thomas Jech. The original statement is: Let $\lambda \ge \kappa$ and let $U$ be a normal measure on $\mathcal{P}_\kappa({\lambda})$. The ...
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3answers
201 views

Beth cardinals and inacceesible cardinals

Since my last question here about the Alephs was too imprecise and thus went over like a lead balloon, I am trying a new and simpler question which asks what I probably should have asked before. ...
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113 views

$(\kappa,\lambda)$-extenders and well-foundedness

(Note that I'm considering extenders within the full context of $\mathsf{ZFC}$ rather than various fragments.) For $E$ a $(\kappa,\lambda)$-extender, the well-foundedness of the limit ultrapower $\...
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1answer
89 views

Can $\mathrm{HOD}^{L(\mathbb R)}$ change between models with large cardinals?

It is a celebrated result that if there is a proper class of Woodin cardinals, then the theory $L(\mathbb R)$ cannot be changed by set-sized forcing. Assume there is a proper class of Woodins. Let $\...
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1answer
87 views

The first uncountable ordinal and some stationary sets

We define a subset $A$ of an ordinal $\alpha$ as stationary iff we have $A$ intersecting every closed and unbounded subset of $\alpha$. Equivalently, one can define this as $$\forall f:\alpha\mapsto\...
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1answer
115 views

Is being a Reinhardt cardinal first-order definable?

As is well known, Reinhardt cardinals are inconsistent with $\mathsf{ZFC}$, but many of the proofs I've seen of this rely on combinatorial or club/stationary set properties. If there's a (somewhat ...
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1answer
135 views

Exploiting Grothendieck universes

A Grothendieck universe provides an easy-to-understand example for a model of ZFC. Because ZFC, if consistent, cannot prove the existence of any model of itself, the existence of universes needs to be ...
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1answer
142 views

Proof of every measurable cardinal carries a normal measure

I'm reading the proof of Theorem 10.20 in Set Theory by Jech and I don't understand the last argument. The theorem says every measurable cardinal carries a normal measure. The proof goes: Let $U$ be ...
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1answer
75 views

characterization of a normal measure

Problem 10.5 from Set Theory, Jech: A measure $U$ on $\kappa$ is normal if and only if the diagonal function $d(\alpha)=\alpha$ is the least function $f$ with the property that for all $\gamma<\...
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2answers
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Do inaccessible-sized sets have any properties that smaller sets don't?

When going from finite to infinite sets, properties arise that are impossible for finite sets. For example, an infinite set has bijections to proper subsets, and to the Cartesian product of the set ...

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