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 ...
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 ...
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 ...
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&...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 $...
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 ...
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 ...
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 ...
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^{\...
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. $\text{ZF}+\text{GCH}\vdash \text{AC}$. ...
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....
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 ...
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 ...