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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.
6
votes
Accepted
Do models of ZFC have arbitrarily large descent values?
No, the descent value is always at most $1$ in any model of ZFC, whether it is well-founded or not. To see this, observe that if the descent value isn't $0$, then on the next step you have $X=L_{\omeg …
7
votes
Accepted
Kunen's inconsistency concerning $L$
Your way of describing the Kunen inconsistency is inaccurate in several respects.
First, the Kunen inconsistency isn't about embeddings of $V$ onto $V$, in the sense of surjective maps, since it is …
6
votes
Are there fragments of set theory which are axiomatized with only bounded (restricted) quant...
There are a few things to say.
First, as Goldstern notes in the comments, there are strictly
speaking no sentences in the language of set theory having only
bounded quantifiers (a sentence is a well- …
5
votes
Which axioms of ZF are used for finite choice?
As mentioned in the comments, one needs truly very few ZF axioms to prove the instances of of the axioms of choice for finite families. Let us denote by $\text{AC}\upharpoonright n$ the assertion that …
9
votes
What is the maximal number of distinct values of the product of n permuted ordinals
For a lower-bound partial answer, note that there are at least as many ways to form a product as a sum, and so $F(n)\leq G(n)$. This is because we may replace an ordinal $\alpha$ with $\omega^\alpha$ …
4
votes
Accepted
Injection of every proper class in the ordinal class
This is equivalent to global choice, since if $V$ itself injects into Ord, then there is a global well-ordering (defined by $x<y$ if $x$ maps to a smaller ordinal than $y$), and vice versa.
9
votes
Accepted
Injection of the proper class of ordinals in every proper class
The answer is no. That principle is equivalent to global choice.
To see this, consider the class $W$ consisting of all
well-orderings of any rank-initial segment $V_\alpha$, for any
$\alpha$. If we h …
1
vote
Order in bijective-equivalent collections of proper classes in set-theory
With regard to your first question, I claim that global choice is equivalent to the assertion that any two classes are comparable under injectivity.
If global choice holds, then this is clearly the …
5
votes
Accepted
A question on models of set theory and Lebesgue measure
This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so …
7
votes
The 'class version' of almost disjoint sets: can it fail?
Here is a way to formalize the almost-disjointness phenomenon in GBC. You don't need an
inaccessible cardinal.
Theorem. In any model of GBC, there is definable
transformation of classes $X\mapsto X^* …
11
votes
Accepted
Sharps and Every Set is Constructible from a Real
The theory is not consistent, since $\mathbb{R}$ is a set, but if $\mathbb{R}\in L[x]$, then every real is in $L[x]$, and this contradicts the existence of $x^\sharp$.
5
votes
Accepted
Questions about a possible way of representing construcive ordinal numbers
The answer to the first question is No and the second question Yes, because in fact every countable (successor) ordinal arises that way. (The successor part is only because you insisted that $L$ has a …
9
votes
Are Sharps in Countable Models Really Sharps?
The answer is no, not necessarily. Countable transitive models can have fake sharps.
To see this, observe simply that the assertion,
$$\hbox{There is a countable transitive model of $\text{ZFC}+x^\ …
9
votes
Accepted
Very weak notions of Universes in ZFC
Unfortunately, I think that this universe concept will not be very useful, since although you add all functions from $a$ to $b$, you haven't ensured that the elements of $a$ or $b$ are in the universe …
10
votes
Accepted
Definitional complexity of truth in $L$ without $0^{\#}$
The answer is yes, the theory of $L$ can be definable by a low-complexity definition quantifying over reals, even when $0^\sharp$ does not exist.
Here is one way to achieve this. Let me assume that …