The axiom-of-choice tag has no usage guidance.

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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**

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**0**answers

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**

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**0**answers

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**

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**0**answers

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**

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**0**answers

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**

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**0**answers

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**

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**0**answers

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**

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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**

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**0**answers

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**

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**0**answers

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**

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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**

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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**

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**0**answers

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**

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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**

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**0**answers

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**

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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**

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**0**answers

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 ...

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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**

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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**

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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**

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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 ...

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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 ...

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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**

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**0**answers

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 ...

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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 ...

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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**

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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 ...

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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 ...

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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**

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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 ...

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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 ...

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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 ...

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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 ...

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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**

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**0**answers

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**

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**0**answers

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 ...

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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**

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190 views

### Logical relationships between weakenings of AC

What are the known logical implications between weak choice principles like $DC_\kappa$", the ...

**1**

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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, ...