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4
votes
2answers
281 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 ...
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
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
1answer
229 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 ...
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
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
2answers
173 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 ...
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 ...
21
votes
2answers
1k views

Axiom of Choice: Ultrafilter vs. Vitali set

It is well known that from a free (non-principal) ultrafilter on $\omega$ one can define a non-measurable set of reals. The older example of a non-measurable set is the Vitali set, a set of ...
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, ...
5
votes
1answer
356 views

Comparability implies well-orderability?

I am trying to prove a small proposition that got me completely stumped, and I cannot find a single counterexample. (ZF) Suppose that $E$ is such that for every $A\subseteq\mathcal P(E)$ either ...
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 ...
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
228 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
210 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
564 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 ...
27
votes
4answers
7k views

Non Borel sets without axiom of choice

This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...
7
votes
0answers
140 views

Strength of claims about extensions of partial preorders and orders to linear ones

Consider these two axioms: Every partial order extends to a linear order. Every partial preorder (reflexive and transitive relation) extends to a linear preorder while preserving strict orderings: ...
5
votes
3answers
226 views

Well-ordering with a topological property

Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it. An ...
2
votes
0answers
345 views

Can we prove an open affine subscheme of a noetherian scheme is noetherian without Axiom of Choice?

I'm interested in proving basic results of algebraic geometry without Axiom of Choice. As for why I think this is interesting, please see Pete L. Clark's answer to this question. To state my problem, ...
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 ...
4
votes
2answers
362 views

Properties of vector spaces without AC

With AC, it is easy to see that any vector space is injective, and free, therefore alse flat and projective. Without AC, vector spaces can be not free. Are they must be projective modules? Flat ...
3
votes
1answer
742 views

Can we construct cohomolgy theory on noetherian separated schemes without Axiom of Choice?

The usual cohomology theory on schemes uses injective or flasque resolutions of quasi-coherent sheaves. Hence it uses Axiom of Choice. However, if the base scheme is a noetherian separated scheme, the ...
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 ...
15
votes
0answers
300 views

Are the reals really a fraction field?

In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$. A crucial point in the ...
5
votes
4answers
234 views

Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?

This question comes after the comments in the recent related question Sigma-complete Lindenbaum algebras?, but in its current form is sufficiently different in my opinion, and so I decided to follow ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
13
votes
1answer
668 views

Sizes of bases of vector spaces without the axiom of choice

Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two ...
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 ...
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 ...
6
votes
0answers
272 views

Can we prove that the ring of formal power series over a noetherian ring is noetherian without axiom of choice?

Let $A$ be a commutative ring with an identity. Suppose that every non-empty set of ideals of $A$ has a maximal element. Let $A[[x]]$ be the formal power series ring over $A$. Can we prove that every ...
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 ...
17
votes
3answers
1k views

Probabilities in a riddle involving axiom of choice

The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question): The Riddle: We assume ...
1
vote
1answer
131 views

Group morphism and axiom of choice

Let $n$ be a strictly positive natural integer. Let us consider the topological group $(\mathbb{R}^n,+)$ with its usual structure. In ZF, can we deduce some form of the axiom of choice from the ...
10
votes
1answer
708 views

Can an infinite number of mathematicians guess the number in a box with only one error?

In this question the following observation was made: Consider a sequence of boxes numbered 0, 1, ... each containing one real number. The real number cannot be seen unless the box is opened. Define ...
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 ...
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. $\text{ZF}+\text{GCH}\vdash \text{AC}$. ...
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 ...
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 ...
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 ...