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19
votes
2answers
2k views

Can a Vitali set be Lebesgue measurable? (ZF)

Here is the definition of Lebesgue measure. The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In ...
29
votes
0answers
1k views

Concerning the various proofs from the axiom of choice that R^3 admits of surprising geometrical decompositions into circles, skew lines and so on: can we prove in any instance that there are no Borel such decompositions? Or that AC is required?

This question follows up on a comment I made on Joseph O'Rourke's recent question, one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of ...
21
votes
0answers
857 views

Supercompact and Reinhardt cardinals without choice

A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it: Definition. A cardinal $\kappa$ is supercompact if for all ordinals ...
20
votes
0answers
1k views

Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice

While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then ...
17
votes
0answers
578 views

Antichains of Cardinals in ZF Without Choice…

With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what ...
16
votes
0answers
345 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 ...
12
votes
0answers
370 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 ...
10
votes
0answers
244 views

Is the universality of the surreal number line a weak global choice principle?

I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear ...
9
votes
0answers
181 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 ...
9
votes
0answers
835 views

Does every group of order bigger than 2 have a non-trivial automorphism?

If $G$ is a non-abelian group, then it has a non-trivial inner automorphism (conjugation by any non-central element). If $G$ is abelian of exponent bigger than 2, then the inversion map is an ...
8
votes
0answers
267 views

A Banach-Tarski game

This is partially inspired by the question http://math.stackexchange.com/questions/1383397/cutting-a-banach-tarski-cake, which I find intriguing if unclearly written. A paradoxical family of subsets ...
8
votes
0answers
539 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 ...
8
votes
0answers
473 views

Seemingly elementary geometric problem in R^3 which requires the axiom of choice

While playing with what I called "quantum matching", the following problem arose: construct a map $F$ from the unit sphere $S_2$ in $R^3$ to itself such that $F(X)$ is orthogonal to $X$ plus has one ...
8
votes
0answers
572 views

In ZF, when is a disjoint union of metrizable spaces metrizable?

It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of metric spaces is metrizable, simply by rescaling or chopping off the individual metrics to have diameter at most one, and ...
7
votes
0answers
195 views

Countable choice in $L(\mathbb{R}^*_G)$

Let $\lambda$ be a singular strong limit cardinal and let $G \subset \text{Col}(\omega,\mathord{<}\lambda)$ be a $V$-generic filter. Let $\mathbb{R}^*_G = \bigcup_{\alpha < \lambda} ...
7
votes
0answers
161 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: ...
6
votes
0answers
207 views

Axiom of choice and a set in the plane that intersects every line in two points

In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects ...
6
votes
0answers
242 views

Finding non convex functions satisfying a weak form of convexity, without the axiom of choice

If a real-valued function $f$ over reals satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other ...
6
votes
0answers
257 views

A new cardinality living in every forcing extension?

This question is motivated by the papers http://arxiv.org/abs/1405.7456 and http://arxiv.org/abs/1410.1224. Say that a set $X$ is "generically presentable" over $V\models ZF$ if there is some ...
6
votes
0answers
247 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," ...
6
votes
0answers
271 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
0answers
270 views

Cardinal characteristics without choice

(I'm taking my definition of a cardinal characteristic from Blass' excellent article http://www.math.lsa.umich.edu/~ablass/need.pdf, which cites Vojtas/Fremlin/Miller; theirs is more general, but I'm ...
5
votes
0answers
208 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 ...
5
votes
0answers
398 views

Existence of Non-Borel sets in models of “All sets measurable”

We know that the consistency of ZFC+"Exists an inaccessible cardinal" implies the consistency of ZF+DC+"All sets are Lebesgue measurable"; and DC proves the existence of non-Borel sets. J. Truss ...
5
votes
0answers
470 views

Kaplansky's theorem and Axiom of choice

Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow: Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...
4
votes
0answers
164 views

On the Axiom of Choice for Conglomerates and Skeletons

Say that $\mathcal{X}$ is a conglomerate if $\mathcal{X} = \{X_i: i \in I\}$, where each $X_i$ and $I$ are classes. The Axiom of Choice for Conglomerates is the statement: Whenever $\mathcal{X}$ and ...
4
votes
0answers
244 views

Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$

Is it consistent with $\mathsf{ZF}$ (without $\mathsf{AC}$) that there is an infinite set $X$ and a subset $S\subseteq\mathcal P(X)$ of the same cardinality as $\mathcal P(X)$ with the property that ...
4
votes
0answers
148 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 ...
3
votes
0answers
77 views

Does Łoś's theorem imply choice given a free ultrafilter?

In the paper "Łoś's theorem and the boolean prime ideal theorem imply axiom of choice" Howard has shown that Łoś's theorem and the boolean prime ideal theorem imply axiom of choice. At the end of the ...
3
votes
0answers
289 views

How close to being well-orderable does this make my powerset?

Let's work in a set theory without assuming AC (for instance, but not necessarily, ZF). Fix a set $k$ satisfying $k\times k \simeq k$, and consider its powerset $X = 2^k$. I have a technical condition ...
3
votes
0answers
216 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 ...
3
votes
0answers
273 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 ...
3
votes
0answers
346 views

New Foundations and weaker forms of choice

New Foundations (introduced by Quine) proves that $AC$ is false. Out of curiosity, is $NF$ consistent with countable choice or dependent choice? What's the strongest consequence of choice still ...
2
votes
0answers
117 views

Defining Global Choice in terms of strong limit cardinals over $ZF$

In his answer to user33038's mathoverflow question "What axioms are stronger than the Axiom of choice?", Prof. Hamkins writes: "What's more, the axiom of choice is equivalent over $ZF$ to the ...
2
votes
0answers
438 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, ...
2
votes
0answers
79 views

Metric space has a basis countably locally finite

it is know that all metric space has a basis countably locally finite and this result is proved by using axiom of choice. Then, the natural question is: is possible to prove this result without using ...
2
votes
0answers
271 views

on the Axiom of Choice and the Spectrum of Rings

consider the following theorem, when $R$ is a commutative ring with a non-zero identity: A ring $R$ is zero-dimensional if and only if $\mbox{Spec(R)}$ is Hausdorff. The proof uses the Axiom of ...
2
votes
0answers
190 views

Logical relationships between weakenings of AC

What are the known logical implications between weak choice principles like $DC_\kappa$", the ...
1
vote
0answers
148 views

Is tree property inconsistent with Berkeley cardinals in the absence of Axiom of Choice?

On one hand due to Kunen's inconsistency theorem it is known that within $\sf ZF$, large cardinal axioms beyond Reinhardt cardinal are inconsistent with $\sf AC$. Also some recent results of Bagaria, ...