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3
votes
Accepted
almost huge embeddings and stationary correctness
I think the answer is no. If $\kappa$ is huge, and we derive an almost-huge tower from the huge embedding, then the embedding computed from the tower has stationary correctness.
Suppose $h : V \to M …
4
votes
Elementary embeddings with the same critical point
This isn't really an answer, but here's a way you can get the maps as you want with the embeddings defined outside of the model. Assume $0^\sharp$ exists, and let $\langle \alpha_i : i < \omega^2 \ra …
5
votes
Accepted
Definition of ineffability behind reflection principles in set theory
Referring to proper classes by means of constants clearly doesn’t use the “internal properties” of the universe of sets. You’re literally reaching outside the universe and adding an artificial means …
5
votes
Accepted
$\aleph_1$-complete fine measures on $P_\kappa(\lambda)$
Yes, if $F$ is large enough. Let $j : V \to M$ be the embedding derived from an $\aleph_1$-strongly compact ultrafilter $U$ on $P_\kappa(\lambda)$. Let $[\mathrm{id}]$ be the set in $M$ represented …
4
votes
'stationary' almost disjoint families
First let us show that consistently there is no such bound, along with $2^\kappa$ larger than any prescribed cardinal. Assume $\diamondsuit_\kappa$. This is consistent with any large cardinal assumpt …
6
votes
What is known about the least cardinal where $\kappa$ fails to be supercompact?
I'm not sure if this gets to the heart of the question, but here are some observations.
For a successor ordinal $\alpha$, let $\kappa_\alpha$ be the least $\kappa>\alpha$ which is $\kappa^{+\alpha}$-s …
14
votes
Why believe in the existence of large cardinals rather than just their consistency?
I think it can be reasonably argued that the large cardinal notions follow a common conceptual pattern, even a semi-formal template. Once we understood the characterization of measurable cardinals in …
7
votes
capturing small sets in small factors
We now have a full answer. The capturing property is equivalent to weak compactness.
Yair already covered the non-inaccessible case. Now we show that this property implies the tree property at $\ka …
6
votes
Reference for proof that consistency of $\omega_1$-Erdos cardinal implies Con(Chang's Conjec...
A sketch of Silver's original argument appears in section 19 here: http://math.bu.edu/people/aki/e.pdf
5
votes
Accepted
Can Ackermann set theory find a natural interpretation in light\heavy class dichotomy?
This question is more subtle than I originally thought. The answer is that the theory is consistent assuming some large cardinal hypothesis, the existence of $0^\sharp$.
If $0^\sharp$ exists, then t …
3
votes
Preservation of measurable cardinals in mild extensions
This is an instance of a more general phenomenon. Let $\kappa$ be a regular cardinal, and $I$ be a $\kappa$-complete ideal on some set $Z$. Let $\mathbb{P}$ be a forcing of size less than $\kappa$, …
6
votes
Accepted
Large cardinals and measurability in $L(A)$
For each $\alpha<\omega_1$, choose a real $r_\alpha$ that codes the ordertype $\alpha$. This sequence $\langle r_\alpha : \alpha < \omega_1 \rangle$ then codes a sequence of surjections from $\omega$ …
3
votes
Accepted
Can a generic ultrafilter over $\mathrm{NS}^+_{\omega_1}$ witness $\omega_1$ is Ramsey-like?
The statement implies that for any $f : [\omega_1]^2 \to 2$, there is an uncountable $S$ such that $f$ is constant on $[S]^2$. This is impossible since $\omega_1$ is not weakly compact. See Jech, Le …
9
votes
Accepted
Why isn't there more interest in "large powerset axioms"?
I suppose there is some interest in propositions implying large powerset. For example, a classical question at the infancy of set theory and real analysis was whether there is a probability measure on …
7
votes
Can we have a $\kappa$-Suslin tree where $\kappa$ is above a measurable cardinal?
To complement Joel's answer and tie into the issue of $L$, consider a model $L[U]$, where $U$ is a normal measure on a cardinal $\kappa$. It is well-known that $L[U]$ satisfies GCH and $\diamondsuit_ …