All Questions
19 questions
18
votes
1
answer
554
views
When can we add choice to a model of ZF
For countable transitive models of ZF, is existence of a ZFC extension with the same height a first order property?
In other words, is there a statement $τ$ (in the language of set theory) such that ...
- 7,545
16
votes
1
answer
1k
views
Can there be a global linear ordering of the universe without a global well-ordering of the universe?
This question arose in the answers to Asaf Karagila's
question Does ZFC prove that the universe is linearly orderable?. The answer there was that one can have a ZFC model with no global linear ...
- 236k
13
votes
3
answers
796
views
How to make countably closed forcing "nice" without choice
When working over a model $V$ of $ZFC$, countably closed forcings are extremely nice:
If $\mathbb{P}$ is countably closed, then $V[G]$ has no new $\omega$-sequences of elements of $V$. In ...
- 20.5k
12
votes
1
answer
695
views
A new cardinality living in every forcing extension?
I'm broadly interested in notions of "generic presentability" - when a given object exists in every forcing extension of the universe by some fixed forcing, at least up to the appropriate ...
- 20.5k
10
votes
4
answers
554
views
What are some kinds of models where DC holds?
There are a lot of ways to build a model where DC fails. However, all of them that I'm aware of involve adding at least a messy set of reals (or rather, taking a forcing extension and then passing to ...
- 20.5k
10
votes
1
answer
761
views
Forcing, cuts, and Dedekind-finite cardinalities
Tl;dr version: there are two natural classes of cuts in the nonstandard model of arithmetic consisting of the Dedekind-finite sets (if, in fact, they constitute such a model); both these classes are ...
- 20.5k
7
votes
2
answers
460
views
Possible Choices for Cofinality of $\aleph_n$ without Choice
$\text{ZFC}$ proves that each $\aleph_{n}$ for $n\in \omega$ is a regular cardinal. But it seems without the Axiom of Choice there are many consistent possible choices for cofinality of such cardinals....
user42090
7
votes
1
answer
401
views
How hard is it to get "absolutely" no amorphous sets?
A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets&...
- 20.5k
6
votes
1
answer
337
views
Intermediate submodels which do not satisfy AC
The following is known:
Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of $...
- 32.1k
6
votes
1
answer
546
views
On successive regular cardinals with no ladders
Definition: Let $\kappa$ be an $\aleph$ cardinal, we say that $\langle f_\alpha\colon\alpha\to\kappa\mid\alpha<\kappa^+\rangle$ is a ladder if every $f_\alpha$ is injective.
Equivalently this is ...
- 39.8k
6
votes
0
answers
300
views
What are these sets in Freyd's model?
Recall Freyd's model of $ZF +\neg AC$ (as recounted in MacLane and Moerdijk's book Sheaves in Geometry and Logic): it arises as the Fourman interpretation of the topos of sheaves on a particular ...
- 35.5k
5
votes
2
answers
610
views
Forcing the nonexistence of a certain set
I have a certain set-theoretic axiom (WISC) which follows from Choice (this is a nuking a fly BTW), but which I suspect is independent of ZF. To show this I need to show that a certain set does not ...
- 35.5k
5
votes
0
answers
146
views
$2^{|V|}$ class cardinalities without global choice
Is it consistent with Morse-Kelley set theory without global choice (but with choice for sets) that there are $2^{|V|}$ proper classes of different cardinalities?
Alternative question: Is it ...
- 7,545
4
votes
2
answers
753
views
Minimal Generalized Continuum Hypothesis & Axiom of Choice
It is well known that working in the frame of $\text{ZF}$, the Generalized Continuum Hypothesis ($\text{GCH}$) implies the Axiom of Choice ($\text{AC}$), i.e. $\text{ZF}+\text{GCH}\vdash \text{AC}$.
...
user42090
4
votes
1
answer
219
views
Generic Absoluteness restricted to formulas with low complexity or to the class of forcings
Ikegami and Schlicht proposed a principle, namely generic absoluteness, which is stated below using Hamkins' and Lowe's terminology:
Working in $ZF$:
(Generic Absoluteness): For all formulas in ...
- 2,381
4
votes
0
answers
166
views
Consistency of definability beyond P(Ord) in ZF
Is it consistent with ZF that the satisfaction relation of $L(P(Ord))$ is $Δ^V_2$ definable? More generally, is it consistent with ZF that there is a $Δ^V_2$ formula (taking $α$ as a parameter) that ...
- 7,545
4
votes
0
answers
379
views
Forcing without choice: when countable sets yield reals
One natural way to show that a forcing adds no new reals is to show that it is countable closed (EDIT: this is somewhat misleading, see Joel's comment below). However, it turns out that this is ...
- 20.5k
2
votes
1
answer
784
views
Subsets of Real Numbers (Edited & Revised Version)
Question 1: Is it consistent with $\text{ZF}$ that only countable subsets of $\mathbb{R}$ are well-orderable?
Question 2: Is it consistent that for some $\lambda$, $\aleph_0 < \lambda < 2^{\...
user42090
1
vote
1
answer
156
views
Invariant names and submodels of forcing extensions
EDIT: There are serious problems with the definition below; see the comment thread below for those problems and some thoughts on addressing them. I'm leaving the question up for now since I think the ...
- 20.5k