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7
votes
Accepted
Property of $L$ Relating to Reflection
The answer to the "More precisely ..." question is that yes, it may be the case that $L$ has a single inaccessible cardinal $\kappa$ and that for no larger $\beta >\kappa$ is there any $\alpha$ with $ …
8
votes
What is the strength of chains of 1-extendibles?
I think for the Theorem one can get by with a subcompact cardinal, which is weaker than a $\kappa^+$-supercompact.
( $\kappa$ is subcompact if for any $A\subseteq H_{\kappa^+}$ there are $j,\mu < \ka …
13
votes
Accepted
Erdős cardinals and ineffable cardinals
To see that the set of ineffables is stationary, let $C\subseteq \eta_\omega$ be any closed and unbounded set. It suffices to show that the good indiscernibles arising in the definition of $\omega$-Er …
5
votes
Relation between indiscernibles for $L$ and for $L[A]$
It can also be the case that $I_A$ is a periodic (but club) subclass of $I$: by Jensen Coding one can define (necessarily by class forcing) reals $a\subset\omega$ with $0^\sharp \notin L[a]$ and that …
23
votes
Accepted
Why does inner model theory need so much descriptive set theory (and vice versa)?
I think that in some ways you have answered your question yourself: we see that to prove properties about sets, say within the projective hierarchy, we need representations of those sets of reals as t …
11
votes
Accepted
Inner model in which every uncountable cardinal is large
Yes: an assumption like the one you quote for inaccessibles in L, namely $0^\sharp$. Instead you take a "mouse" i.e. an iterable structure, which has a measure of Mitchell order 1 as the topmost fina …
10
votes
Accepted
Inner model theory without choice
Addressing the first question: I should argue that Choice comes in almost at the beginning of the inner model project, if we regard proving Covering Lemmata as an integral part of that project: one de …
9
votes
Accepted
Do indiscernibility embeddings exist for an initial segment of an inner model of many measur...
My reading of this question was different from Andreas', because Norman asked for order preserving maps of the indiscernibles to extend to embeddings
$j:L[U]_\theta \rightarrow$ $L[U] _\theta$
i.e …
5
votes
Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0#
The answer to Q2 is 'No'. Suppose $j:L\rightarrow L$ is a non-trivial elementary embedding. We use the following fact:
$\bullet$ $cp(j)$ (the first ordinal moved by $j$) is always a Silver indiscer …
10
votes
Large Cardinal Principles that Imply $\Sigma_3^1$-Generic Absoluteness
A proper class of measurables more than suffices.
It suffices for the generic absoluteness to have X-sharp exists for every set of ordinals X. Then the Martin-Solovay tree can be constructed through …
25
votes
Accepted
Devlin's "Constructibility" as a resource
Mathias has a paper where he corrects the flaws that occur in Devlin's theory BS (= Basic Set Theory). The theory has to be only slightly strengthened to be correct. (It is more than sufficent to add …
13
votes
Accepted
Is Prikry forcing minimal?
There is a theorem of Gitik, Kanovei and Koepke that characterises the degrees of constructibility in $M[G]$ where $G$ is prikry generic over the model $M$: they are isomorphic to the $P(\omega)/Fin$ …
6
votes
Consistency of many Erdos cardinals
I agree with the last sentence of Bill Mitchell's comment. But here is something closer than the cardinals mentioned. In [1] the notion of "almost Ramsey" cardinal was coined. Such a cardinal $\kappa$ …
24
votes
Latest status of core model theory?
I think the short answer is that we are still a long way from a full core model for a supercompact cardinal.
We have core models (in the sense of Steel's `Core Model Iterability Problem' Lecture Note …
10
votes
Why "adding" a single extender cannot give an L-like inner model for say, a strong cardinal?
One point to make is that no one embedding witnesses that a cardinal is strong (that is unless it is witnessing that $\kappa$ is something much stronger, like a supercompact).
The definition requires …