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

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### 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 ...

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### 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 ...

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### 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 ...

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### Some questions regarding Shelah's revised Generalized Continuum Hypothesis [closed]

It is well known that $\mathsf{ZF}+\mathsf{GCH}\vdash\mathsf{AC}$ (which means that the Kunen inconsistency can be proven in $\mathsf{ZF}+\mathsf{GCH}$). Consider now Shelah's revised Generalized ...

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### Does this axiom (a weak form of class valued choice) has a name?

At some point in my work (which has nothing to do with set theoretics foundation) I need to consider the following axiom:
For any set $X$, any class $V$ with a surjective map $f : V ...

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### Does “cardinal arithmetic is well-defined” imply axiom of choice?

Let me quickly explain what I mean with my question.
Let $(\kappa_i)_{i\in I}$ be a collection of cardinal numbers, indexed by elements of some set $I$. We can try to define $\sum\limits_{i\in ...

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### Freeness of the group of principal ideals of a number field

This is just a question wondering whether a concrete (enough) result that can be proved using the axiom of choice can be proved without it. The result being that the group of principal ideals $P_K$ of ...

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### Set theories that do require the existence of urelements?

I am looking for an axiomatic set theory that not only admits the existence of urelements/atoms (via two-sortedness or an additional unary predicate) but requires it, e.g. by an axiom like "for each ...

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### 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 ...

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274 views

### A question about Cantor's Power Set theorem without the Axiom of Choice

Assume that we are working in ZF set theory without the Axiom of Choice. If S is an infinite set, let $S(f)$ denote the set of all finite subsets of $S$, let $S(I)$ denote the set of all infinite ...

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### Axiom of choice for sets of finite sets

The question I am going to ask is really to satisfy my curiosity, as I am not at all an expert of the subject and do not plan to really work on it. Hence, if you think the question is not suitable for ...

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### The patterns of possibility for nontrivial automorphisms and nontrivial elementary embeddings of the universe

In their paper "The Role of the Foundation Axiom in the Kunen Inconsistency" (arXiv:1311.0814 [Math.LO]), Daghighi, Golshani, Hamkins, and Jerabek show that the patterns of possibility for the ...

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### How do I apply the Boolean Prime Ideal Theorem?

I have become aware of an amazing phenomenon from a myriad of questions and answers here on MathOverflow: many of the results that I would typically prove using the Axiom of Choice can actually be ...

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### Why worry about the axiom of choice?

As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory. Yet I still see people fuss about whether or not theorem X depends on it, and I don't ...

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### Logical strength of “choice functions exist for well-ordered families”?

A colleague of mine suggested the following weakening of the axiom of choice:
If $\mathscr{F} := \{F_\alpha\}$ is a well-ordered family of non-empty sets (i.e., there is a bijection between ...

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### The role of the rigid relation principle ($RR$) in the Kunen inconsistency

Consider the rigid relation ($RR$) principle, i.e.
"every set admits a rigid binary relation", that is,"that for every set $A$ there is a binary relation $R$ on $A$ such that the structure $(A,R)$ is ...

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### Notions of infinity in $\mathsf{ZF}$ without choice

Consider the following statements about a given set $X$ in in $\mathsf{ZF}$:
(1) There is $x_0\in X$ such that there is a surjective map $\varphi: X\setminus\{x_0\}\to X$.
(2) There is an injective ...

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### On surjections, idempotence and axiom of choice

The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF?
(I couldn't find it in the Consequences site so far.)
If $A$ is an infinite set such that $A$ ...

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### Independence of the countable axiom of choice

How does one proove that the Countable axiom of choice is not provable in ZF?Is there any brief proof?Does the Independence of the countable axiom of choice implies the independence of the axiom of ...

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### 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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729 views

### Does ZF prove that topological groups are completely regular?

Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$.
Assume $\{\mathbf{e}\}$ is closed in $\langle ...

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### Noetherianess of a finite module over a noethrian ring without Axiom of Choice

All rings are assumed to be commutative with 1.
We say a module over a ring is strictly noetherian if every non-empty set of submodules has a maximal member. We say a ring is strictly noetherian if it ...

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### Completeness of the club filter without AC

Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the ...

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### 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 ...

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### Matching power series to infinity

As pointed out by Makoto, on this question about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another ...

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### 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 ...

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### Does k(X) have a k-basis for every set X, without AC?

This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.
For any field $k$, the field $k(x)$ of rational functions in one variable has an ...

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### Are two forms of the Dual Schroeder-Bernstein property equivalent?

We know the Shroeder-Bernstein (SB) theorem can be proved in ZF, while the Dual Schroeder-Bernstein (DSB) can be proved in ZF+AC but not in ZF. Define as ISB the property that whenever there are both ...

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### Axiom of choice and vector spaces over a given field

It is my impression that the following question is open:
Does the existence of a basis for every vector space over the field K = the reals having a basis imply the axiom of choice?
I saw an answer ...

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### A question regarding the Hahn-Banach theorem

Wikipedia states that, in $ZF$, the Axiom of Choice ($AC$) implies the Hahn-Banach theorem, but that the Hahn-Banach theorem does not imply $AC$. It also states that in $ZF$, the Hahn-Banach theorem ...

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### What is known about global well ordering of classes in Gödel-Bernays?

I would like to have something like a linear order on classes, such that every instantiated predicate of classes has a minimal instance in that order. For my purposes, it is fine to assume V=L for ...

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### 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 ...

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### 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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### Subset of the plane that intersects every line exactly twice

In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to ...

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### 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 ...

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### When does Skolemization require the axiom of choice?

Skolemization is often used for eliminating existential quantifiers, which is often useful for proving theorems, especially in automated resolution theorem proving. Skolemization in first order ...

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### Compactness of the Hilbert cube without the Axiom of Choice

I am just curious: is there a published proof of the compactness of the Hilbert cube that does not use the Axiom of Choice, or is it well known?

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### Are there known ways to posit definable global choice in ZF without positing V=L?

I need a global choice function defined by a formula in (a fragment of) ZF. There is no harm in assuming V=L for my purposes. But I wonder if there are any familiar alternative ways to get this?
...

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### What are the current views on consistency of Reinhardt cardinals without AC?

It's well known that Reinhardt cardinals are inconsistent, provided that we have access to axiom of choice, but, as far as I know, we are clueless about this when we don't assume choice. For me, the ...

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### Difference between ZFC and ZF+GCH

I hear that the axiom of choice (AC) derives from
The generalized continuum hypothesis(GCH).
And also hear that both AC and GCH are independent of
Zermelo–Fraenkel set theory(ZF).
So, I'm just ...

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### 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} ...

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### Is it possible to formulate the axiom of choice as the existence of a survival strategy?

Consider the following situation:
There is an infinite set $G$ of giraffes.
A lion comes and announces a set $C$ of all possible colours and an infinite cardinal $\kappa$.
The hungry lion ...

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### Categorifications of Zorn's lemma

I'm wondering about categorifications of Zorn's lemma along the following lines.
Lemma: if $\mathbf{C}$ is a small category in which every directed diagram of monomorphisms has a cocone of ...

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### Most 'unintuitive' application of the Axiom of Choice?

It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle, Tychonoff's theorem, and the fact that every vector space has a basis. Even ...

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### Does existence of $\omega_1$ subset of reals imply $\omega_1$ choice for subsets of reals?

Suppose there exists a subset of $\Bbb R$ which has cardinality $\omega_1$. Is it then necessarilly true that for every collection of $\omega_1$ subsets of $\Bbb R$ there exists a choice function?
I ...

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### 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 ...

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### 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 ...

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### Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?

Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial(scaler multiplication is not identicaly zero) $\mathbb{C}$-module.
Now my questions are?
0.Is it consistent with $ZF$ that $\mathbb{R}$ is ...

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### Objects which can't be defined without making choices but which end up independent of the choice

It happens a lot of times that when one defines a new object (ring, module, space, group, algebra, morphism, whatever) out of given data one first chooses some additional structure. And sometimes ...

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### 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 ...