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8
votes
1
answer
269
views
What is the consistency strength of "Singular worldly that is inaccessible in an inner model"?
In short, what can we say about the consistency strength of "$\kappa$ is a singular worldly and inaccessible in an inner model"?
Clearly, $0^\#$ exists since we have a singular cardinal which is regul …
11
votes
1
answer
417
views
Coding the universe into a real over better core models
One of the most incredible results in modern set theory, due to Jensen, is that given any model of $\sf ZFC$, there is a class forcing which adds a real number $r$ and in the extension $V=L[r]$. Moreo …
4
votes
0
answers
174
views
Sequences of sequences of sequences and elementary embeddings
Suppose that $\kappa$ is the critical point of $j\colon V\to M$, and suppose that $\mathcal F=(F_\alpha\mid\alpha\leq\kappa)$ is a sequence such that for every limit ordinal $\alpha$, $F_\alpha$ is a …
6
votes
0
answers
317
views
Temporary destruction of measures in intermediate models
It is a well-known theorem that if $\kappa$ is measurable, then there is a generic extension in which $\kappa$ is no longer weakly compact, but we can force its weak compactness back and recover the f …
6
votes
0
answers
401
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 …
6
votes
1
answer
238
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 c …
13
votes
1
answer
758
views
Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncoun...
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 w …
14
votes
1
answer
444
views
What are some good lower bounds on the consistency of the failure of the PCF conjecture?
Shelah's celebrated theorem states that $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega}<\aleph_{\omega_4}$.
But the conjecture is that $\omega_4$ can be provably replaced by $\ome …
9
votes
2
answers
420
views
"Weakly" Woodin cardinals
Recall that an inaccessible cardinal $\kappa$ is a Woodin cardinal if for every $A\subseteq V_\kappa$ there is an unbounded set in $\kappa$ of $\lambda$ such that $V_\kappa\models\lambda$ is $A$-stron …
11
votes
1
answer
533
views
What is the consistency strength of "Every set is a member of a transitive model"?
Recall that $\kappa$ is a worldly cardinal if $V_\kappa$ is a model of $\sf ZFC$. While every worldly cardinal is a strong limit cardinal, it is not necessarily regular. The point being that the short …
5
votes
0
answers
248
views
Natural combinatorial properties of $\omega_1$ and weakly compact cardinals
One of the magnificent theorems of $\sf ZFC$ is that there exists an Aronszajn tree on $\omega_1$. Namely, a tree of height $\omega_1$ in which every level is countable, but no branch is cofinal.
On …
7
votes
2
answers
1k
views
Embedding property of weakly compact cardinals
One of the characterizations of $\kappa$ being a weakly compact cardinal is being inaccessible, and for every $\kappa$-model $M$, there is a [$\kappa$-model] $N$ and an elementary embedding $j\colon M …
8
votes
1
answer
566
views
What consistency results follow the assumption: $\forall\alpha(\aleph_{\alpha+1}\nleq2^{\ale...
In a recent question on Math.SE it was asked whether or not For every infinite cardinal $\mathfrak m$ there is no $\aleph$ number, $\kappa$, such that $\mathfrak m<\kappa<2^{\mathfrak m}$.
By requiri …
3
votes
1
answer
379
views
How to change the successor of a singular with a Woodin?
I'm looking for references on how to change the successor of a singular cardinal from "more or less" minimal assumptions. If possible, then without adding bounded subsets to the singular either.
In s …
8
votes
1
answer
655
views
Are superstrong stronger than strongly compact cardinals? (or vice versa)
In the last part of Kanamori's excellent "The Higher Infinite" there is a small diagram about the strength and consistency strength of some major large cardinal axioms.
Below supercompact cardinals t …