4
votes
2answers
296 views

Exponentiation and Dedekind-finite cardinals

It is known that the sum and the product of two Dedekind-finite cardinals are also Dedekind-finite cardinals. What about cardinal exponentiation ? Question: Let A and B be two Dedekind-finite ...
5
votes
1answer
226 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 ...
7
votes
0answers
110 views

Without AC, which implications between the different definitions of amenability still hold?

More precisely, I would like to know which implications between the following definitions of amenability of a discrete countable (or even finitely generated) group can be proved to hold with only ZF ...
4
votes
1answer
148 views

Is it compatible with ZF to assume that every amenable discrete group is finite?

The question is in the title, amenability being understood as the existence of a left-invariant finitely additive probability measure on the group of interest. The case of countable groups is treated ...
7
votes
1answer
232 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 ...
13
votes
1answer
846 views

How much of GCH do we need to guarantee well-ordering of continuum?

It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, ...
6
votes
0answers
192 views

Does Sageev's result need an inaccessible?

In 1981, building on work by Ellentuck in 1974, Sageev showed ("A model of ZF + there exists an inaccessible, in which the Dedekind cardinals constitute a natural non-standard model of arithmetic," ...
3
votes
1answer
229 views

What axioms (other than choice) have a taming effect on the ordering of cardinalities?

Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, ...
4
votes
1answer
211 views

Choice principle strong enough to defy $V=L$

When using axiom of choice in proofs, people often say that this is non-constructive because AC gives us only proofs of existence, without giving explicit example. However, because in $L$ AC holds, we ...
21
votes
1answer
565 views

Is there a model of ZF set theory with a set that does not inject into the cardinals?

Question. Is there is a model of ZF set theory with a set $X$ that does not inject into the cardinals? I use the term "cardinal" here in the ZF sense, so they are not necessarily well-orderable. To ...
6
votes
1answer
213 views

Sets of cardinalities of bases without choice

For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...
14
votes
2answers
456 views

Pathological behavior of Borel sets?

Usually in set theory, Borel sets are much more nicely behaved than arbitrary sets of reals. One reason for this is Borel determinacy, which immediately yields measurability, Baireness, and the ...
1
vote
3answers
281 views

Why doesn't choice imply global choice (in NBG)?

I thought ZFC proved the existence of an inductive well-ordering that is itself a set for any stage of V. NBG with only the regular AC should then prove/assert the existence of a class R of ordered ...
17
votes
1answer
253 views

Linear maps between arbitrarily chosen vectors of vector spaces $V$ and $W$

I recently came across this question: Is the axiom of choice needed to prove the following statement: Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. ...
5
votes
1answer
147 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 ...
4
votes
1answer
257 views

Re-interpreting vector spaces in a choice-less model of ZF as modules over a regular ring in ZFC

I am searching a module $M$ over a (von Neumann) regular ring $A$ ($\forall a\in A$, $\exists x\in A$: $axa=a$) with two properties: (1) every finitely generated submodule of $M$ is projective ...
5
votes
1answer
141 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 ...
2
votes
0answers
167 views

How many Dedekind-finite sets can $\mathbb{R}$ be partitioned into?

Building off Asaf Karagila's answer to my previous question (Can $\mathbb{R}$ be partitioned into dedekind-finite sets?) on partitioning $\mathbb{R}$ into strictly Dedekind-finite sets: (1) What ...
11
votes
4answers
836 views

Weak forms of the Axiom of Choice

Let $n\geq 2$ be a natural number and consider the following: $AC(n)$: For each family $\{X_i\}_{i \in I}$ of $n$-element sets the product $\prod_{i\in I}X_i$ is non-empty. Is it known that for ...
8
votes
1answer
242 views

Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...
5
votes
3answers
219 views

In $L$, does there exist a definable non-principal ultrafilter on $\mathbb{N}$

The axiom of constructibility $V=L$ leads to some very interesting consequences, one of which is that it becomes possible to give explicit constructions of some of the "weird" results of AC. For ...
4
votes
0answers
135 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 ...
5
votes
1answer
202 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 ...
2
votes
1answer
561 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 < ...
4
votes
3answers
341 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. $\text{ZF}+\text{GCH}\vdash \text{AC}$. ...
4
votes
1answer
151 views

Discontinuous representations of GL(n,C) in ZF

Discontinuous linear representations of $GL(n,\mathbb{C})$ can be obtained from the so-called "wild" (field) automorphisms of $\mathbb{C}$; but these wild automorphisms in turn require some choice to ...
8
votes
0answers
443 views

Full conditional probabilities and versions of AC?

A probability is a finitely additive measure on a boolean algebra with total measure $1$. A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean ...
4
votes
4answers
392 views

Strength of some claims about finitely additive measures on infinite sets?

Assume ZF. Consider the claim: (1) For any infinite set $\Omega$, there is a finitely additive probability measure $\mu:2^\Omega\to[0,1]$ with $\mu(A) = 0$ whenever $|A|<|\Omega|$. Then (1) is ...
6
votes
2answers
235 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 ...
10
votes
2answers
371 views

Is sigma-additivity of Lebesgue measure deducible from ZF?

Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)? Note 1. ...
6
votes
1answer
195 views

Solovay's Theorem on Partitions of Stationary Sets and Weak Choice Principles

There is a weak choice principle called $DC_\lambda$ which holds in $L(V_{\lambda+1})$ under the assumption of a non-trivial elementary embedding $$j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ and it is ...
2
votes
1answer
206 views

What is the order type of $L$ with Godel's well ordering?

In some sense $Ord$ is a "proper class" ordinal. Unfortunately the notion of a proper class ordinal is not a straight forward generalization of the notion of "set" ordinals because the proper classes ...
5
votes
0answers
163 views

Proving equivalence of a tree-based version of Countable Choice for families of finite sets

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
12
votes
0answers
345 views

How much choice is required to prove concretizability theorems in category theory?

A concretization of a category is a faithful functor to the category of sets. A category is concretizable if there exists such a functor. An evident necessary condition for concretizability is ...
7
votes
2answers
237 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 ...
10
votes
2answers
261 views

Singular successors without large cardinals

Assuming the axiom of choice we have that successor cardinals are regular. However as one of the first examples of uses of forcing show, it is consistent relative to $\sf ZF$ that $\omega_1$ is ...
9
votes
0answers
173 views

The global dimension of fields

In the absence of the Axiom of Choice, it is not necessarily true that all vector spaces over a field have bases. What are the possible global dimensions of fields in a model of ZF in which AoC ...
8
votes
1answer
451 views

A question about the Axiom of Choice

Let AC denote the Axiom of Choice. Let PP denote the so-called "Partition Principle" which states that "If S is a non-empty set and T is a non-empty set of pairwise disjoint subsets of S, then S can ...
5
votes
1answer
202 views

Proper-class sized “ring” with no maximal ideals

Suppose I have a collection of "elements" together with operations that satisfy the axioms for a commutative ring with identity --- except that these elements form not a set, but a proper class. Must ...
4
votes
1answer
183 views

Well-Ordering theorem of cardinal$\kappa$

I've heard from others about the WO($\kappa$) as a counterpart of AC($\kappa$), but I cannot find a suitable way to express it in ZF since "every set of cardnality $\kappa$ can be well-ordered" is ...
9
votes
1answer
298 views

Cardinals without choice: interpolation (reference wanted)

Is there a published reference for this ZF theorem? Let $m,n\in\mathbb{N}$. If $a_1,\dots,a_m$ and $b_1,\dots,b_n$ are cardinals such that $a_i\le b_j$ for all $i$ and $j$, then there is a cardinal ...
8
votes
1answer
378 views

Weakest choice principle required for Robertson-Seymour Graph Minor Theorem?

The main Robertson-Seymour Theorem states that finite graphs form a well-quasi-ordering under the graph minor relation. In other words, in every infinite set of finite graphs, there exist two graphs ...
21
votes
2answers
1k views

Hahn's Embedding Theorem and the oldest open question in set theory

Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his paper Über die nichtarchimedischen Grössensysteme (Sitzungsberichte der ...
2
votes
0answers
294 views

Is the axiom of choice really related to choice? [closed]

I am not an an expert in set theory, so this question could be trivial. I am sorry in that case. Let $I$ be a set and $\{ X_i \}_{i \in I}$ be a collection of sets such that $X_i \neq \emptyset$ for ...
3
votes
1answer
344 views

How much of ZFC do I need to construct this cofinal, order-preserving class function?

EDIT: I'm bumping this, because while Joel ruled out some naive options, my question in bold below is not yet answered. Suppose I have a directed partially ordered set $(\Gamma,\leq)$ with a bottom ...
36
votes
1answer
1k views

When does $A^A=2^A$ without the axiom of choice?

Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$ However without the axiom of choice this doesn't ...
4
votes
2answers
612 views

Are all models of ZF + DC + “All set of reals are lebesgue measurable” also models of CH? [duplicate]

Possible Duplicate: Lebesgue Measurability and Weak CH I have studied a little set theory and I found that Solovay constructed a model of ZF+DC+"All set of reals are Lebesgue measurable" and ...
3
votes
3answers
453 views

Set theory question

It is known from a result of Sierpinski that the generalized continuum hypothesis (GCH) implies the axiom of choice (AC). It is also known from the celebrated results of Cohen that AC is independent ...
11
votes
3answers
1k views

Cantor's diagonal argument and ZF

Cantor's diagonalization construction, on a certain view, furnishes functions $$d_X:{\rm Injections}(X,P(X))\rightarrow P(X)$$ that satisfy $\forall X\forall i\ \ d_X(i)\not\in i(X)$ In ZF, can ...
8
votes
3answers
776 views

Axiom of Choice and continuous functions

Do you know if the following statement is an equivalent form of the axiom of choice or not? If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is ...