Skip to main content

All Questions

Filter by
Sorted by
Tagged with
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 ...
Dmytro Taranovsky's user avatar
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 ...
Dmytro Taranovsky's user avatar
4 votes
2 answers
222 views

Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\lambda$?

Let $\lambda<\kappa$ be cardinals and consider the forcing $\operatorname{Col}(\lambda,\kappa)$ adding a generic surjection $\lambda\to\kappa$. More formally, $\operatorname{Col}(\lambda,\kappa)$ ...
Hanul Jeon's user avatar
  • 3,042
10 votes
1 answer
683 views

Small Violations of Choice: Can we force AC without collapsing the cardinalities of ordinals?

We say that a model $M$ of $\mathsf{ZF}$ satisfies Small Violations of Choice ($\mathsf{SVC}$) if all (any) of the following apply: There is a model $V\subseteq M$ such that $V\vDash\mathsf{ZFC}$, ...
Calliope Ryan-Smith's user avatar
12 votes
1 answer
442 views

Is there a more modern account of the main results of "Adding Dependent Choice" by D. Pincus?

I would like to read Pincus' article Adding dependent choice, where he proves, among other things, the consistency of the theory $\mathsf{ZF+DC+O+\neg AC}$, where $\mathsf{DC}$ stands for Dependent ...
Lorenzo's user avatar
  • 2,286
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&...
Noah Schweber's user avatar
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 ...
Dmytro Taranovsky's user avatar
4 votes
0 answers
142 views

Consistency of a strange (choice-wise) set of reals, pt. 2

This is a follow-up on this question. Consider a set $X\subseteq \mathbb{R}$ such that $X$ is not separable wrt its subspace topology Every countable family of non-empty pairwise disjoint subsets of $...
Lorenzo's user avatar
  • 2,286
7 votes
2 answers
722 views

Consistency of a strange (choice-wise) set of reals

Consider a set $X\subseteq \mathbb{R}$ such that $X$ is not separable wrt its subspace topology For all $r\in\mathbb{R}$ there exists a sequence $(x_n)_{n\in\omega} \subset X$ converging to $r$ In a ...
Lorenzo's user avatar
  • 2,286
7 votes
2 answers
683 views

Can second-order logic identify "amorphous satisfiability"?

Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can ...
Noah Schweber's user avatar
9 votes
1 answer
313 views

Does ZF + BPI alone prove the equivalence between "Baire theorem for compact Hausdorff spaces" and "Rasiowa-Sikorski Lemma for Forcing Posets"?

Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o. $\mathbb{P}$ (i.e. $\mathbb{P}$ is a reflexive transitive relation) and for any countable family of dense subsets of $\mathbb{...
Samuel G. Silva's user avatar
5 votes
1 answer
149 views

Maximality principle in symmetric extensions

Let $M$ be a ctm and $P\in M$ a forcing order. In regular forcing extensions, we have the following well-known Principle: $$p\Vdash_{M,P}\exists x\phi[x]\;\Longrightarrow\;\exists\sigma\in M^P\;p\...
Hannes Jakob's user avatar
  • 1,799
10 votes
1 answer
535 views

Models of ZF intermediate between a model of ZFC and a generic extension

Let $M$ be a countable model of $ZFC$ and $M[G]$ be a (set) generic extension of $M$. Suppose $N$ is a countable model of $ZF$ with $$M\subseteq N \subseteq M[G]$$ and that $N=M(x)$ for some $x\in ...
Toby Meadows's user avatar
  • 1,142
19 votes
2 answers
856 views

Do choice principles in all generic extensions imply AC in $V$?

It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a ...
Elliot Glazer's user avatar
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 ...
Noah Schweber's user avatar
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 ...
Noah Schweber's user avatar
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 ...
Rahman. M's user avatar
  • 2,381
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 ...
Noah Schweber's user avatar
2 votes
1 answer
438 views

The patterns of possibility for nontrivial automorphisms and nontrivial elementary embeddings of the universe

In their paper "The Role of the Foundation Axiom in the Kunen Inconsistency" (arXiv:1311.0814 [Math.LO]), Daghighi, Golshani, Hamkins, and Jerabek show that the patterns of possibility for the ...
Thomas Benjamin's user avatar
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 ...
Noah Schweber's user avatar
3 votes
1 answer
600 views

Does existence of $\omega_1$ subset of reals imply $\omega_1$ choice for subsets of reals?

Suppose there exists a subset of $\Bbb R$ which has cardinality $\omega_1$. Is it then necessarilly true that for every collection of $\omega_1$ subsets of $\Bbb R$ there exists a choice function? I ...
Wojowu's user avatar
  • 28.2k
9 votes
2 answers
1k views

Relationship between fragments of the axiom of choice and the dependent choice principles

The dependent choice principle ${\rm DC}_\kappa$ states that if $S$ is a nonempty set and $R$ is a binary relation such that for every $s\in S^{\lt\kappa}$, there is $x\in S$ with $sRx$, then there ...
Victoria Gitman's user avatar
7 votes
2 answers
349 views

Forcing $\neg AC$

Sorry if this sounds like a silly reference request, but I wasn't able to track down any. I'm looking for proof, via forcing, that axiom of choice can fail in a model of $ZF$. All of papers I found ...
Wojowu's user avatar
  • 28.2k
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 ...
Noah Schweber's user avatar
6 votes
1 answer
171 views

Function Approximation in c.c.c Forcings without AC in Ground Model

Consider the following basic theorem. Theorem. If $M$ is a c.t.m of ZFC and $\mathbb{P}$ a c.c.c forcing notion in $M$ and $G$ a $\mathbb{P}$ - generic filter on $M$ then for all $A,B$ in $M$ and for ...
user avatar
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 $...
Mohammad Golshani's user avatar
4 votes
0 answers
161 views

What is the meaning of restricting a Boolean value to a subalgebra?

$\require{AMScd}$ I am studying the proof that the ordering principle does not imply the axiom of choice in Jech's book "The Axiom of Choice" (Section 5.5). Let $P$ be the set of finite partial ...
Carlos's user avatar
  • 1,688
6 votes
1 answer
696 views

A question about the first Cohen model

Consider the first Cohen model, i.e. let $M$ be a countable transitive model of ZFC + $V=L$, let $\mathbb P$ be the poset consisting of finite partial functions from $\omega\times\omega$ to $2$, let $...
Carlos's user avatar
  • 1,688
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^{\...
user avatar
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. $\text{ZF}+\text{GCH}\vdash \text{AC}$. ...
user avatar
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....
user avatar
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 ...
Noah Schweber's user avatar
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 ...
Joel David Hamkins's user avatar
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 ...
Asaf Karagila's user avatar
  • 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 ...
David Roberts's user avatar
  • 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 ...
David Roberts's user avatar
  • 35.5k
4 votes
1 answer
441 views

Is the ordering principle preserved in generic extensions?

The ordering principle says that every set can be linearly ordered. In a previous question Why are some axioms preserved in generic extensions? Asaf Karagila asked which axioms are preserved in ...
Stefan Geschke's user avatar
6 votes
2 answers
886 views

Why are some axioms preserved in generic extensions?

It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$. On the other hand, it is long known that ...
Asaf Karagila's user avatar
  • 39.8k
18 votes
5 answers
2k views

Forcing over models without the axiom of choice

In the vast majority of papers forcing is always developed over ZFC. Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain ...
Asaf Karagila's user avatar
  • 39.8k
7 votes
1 answer
908 views

Symmetric extensions and class forcing

Suppose $V\models ZFC$ and $P\in V$ is a poset of forcing conditions. It is a basic theorem in forcing that $V[G]\models ZFC$ for any generic extension by a $V$-generic filter $G$. It is also known ...
Asaf Karagila's user avatar
  • 39.8k