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6
votes
Accepted
Constructible cardinality downslides and their consistency strengths?
Since $\omega_1^V\subseteq L$ (simply because $L$'s construction goes through all the ordinals), we obviously can't have $L$ be countable. On the other hand, if $0^\sharp$ exists then every uncountabl …
15
votes
Accepted
What is the least inaccessible cardinal for Tarski-Grothendieck set theory?
I missed the inaccessibility requirement initially - fixed!
Since $\mathsf{TG}$ is "just" $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals," an inaccessible $\alpha$ satisfies $V_\a …
18
votes
A Löwenheim–Skolem–Tarski-like property
Here's a counterexample for $\kappa=\aleph_1$: let $B$ be the structure with underlying set $\mathbb{N}\sqcup\mathcal{P}(\mathbb{N})$, equipped with the usual ordering on $\mathbb{N}$ as well as the $ …
9
votes
Operations on the set of large cardinal axioms
At least for Mahlo-ness, things are pretty simple to describe:
Given a property $P(\kappa)$ of cardinals implying strong inaccessibility, let $P^{\mathrm{Mahlo}}(\kappa)$ be the property "$\kappa$ ha …
4
votes
Would loss of downward absoluteness for large cardinals repeat itself upwardly?
It depends on the large cardinal property.
For example, inaccessibility is downwards-absolute: if $V\models$ "$\kappa$ is inaccessible" and $W$ is an inner model of $V$ then $W\models$ "$\kappa$ is in …
3
votes
Accepted
How large is the supremum of minimal $V$-heights of all first-order set theories formulated ...
Your ordinal $\beta_\mathcal{L}$ is perfectly well-defined: in my opinion it's more easily thought of as $$\sup\{\alpha: \forall \beta<\alpha(V_\beta\not\equiv V_\alpha)\},$$ and this definition shoul …
4
votes
Accepted
What's the consistency status/strength of this limitation principle?
This principle is inconsistent: consider the formula $\theta(x)$ = "$x^+$ is the smallest infinite cardinal at which $\mathsf{CH}$ fails." The formula $\theta$ cannot hold on more than one infinite ca …
4
votes
Accepted
$\mathtt{PSP}$ implies the consistency of inaccessible cardinals
See Proposition 11.5 (and the discussion leading up to it) in Kanamori's book The Higher Infinite. Note that Kanamori states the result a bit more optimally: if $M\models\mathsf{ZF}$ + "$\omega_1$ is …
8
votes
Accepted
Why does the second smallest worldly cardinal believe the smallest worldly cardinal is worldly?
The truth of a first-order sentence $\varphi$ in a structure $\mathfrak{M}$ is absolute between $V$ (= reality) and sufficiently large transitive sets containing $\mathfrak{M}$. In particular, already …
7
votes
Accepted
What sets can be unraveled?
I emailed Itay Neeman, and he told me the following:
As far as I know it's open. I don't think anything is known about
unraveling beyond what you can get from my methods. These give the
Suslin operat …
4
votes
Accepted
Cardinality of infinite towers of Alephs - can tower be more than countable?
"Transfinite towers" often run into the problem of termination: when $F$ is a "nice" function on ordinals (= monotonic, nondecreasing, and continuous), we get $$F(\sup\{F^n(0): n\in\omega\}=\sup\{F^n( …
4
votes
Accepted
smallest ordinal $\alpha$ such that $L \cap P(L_\alpha)$ is uncountable
Really, this was answered in the comments; I'm putting this answer down to move this off the unanswered queue. I've made this CW and will delete it if one of the original commenters adds their own ans …
14
votes
How to understand the interface of the consistency strength hierarchy, reverse mathematics, ...
Sorry this is a bit disjointed - there's a lot of stuff here. I hope this helps though.
All of these notions are applicable in all contexts - or at least, all sufficiently rich contexts (we probabl …
2
votes
Accepted
Is the principle of indifference of hierarchical construction consistent? What's its consist...
This question can be more clearly phrased as:
Is the principle "We can iterate powerset along any (definable) class-well-ordering" (which is really a scheme, appropriately) consistent with ZFC?
…
4
votes
Accepted
Possible inconsistency related to embeddings $j: M\prec V$
There is no contradiction here.
Look at Theorem $2.3$:
Suppose $I\subseteq On$ witnesses $On$ is Ramsey. Then, definably
over $\langle V,\in, I\rangle$, there is a transitive class $M$, and an …