Questions tagged [axiom-of-choice]

An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.

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3 votes
1 answer
163 views

A simple form of choice

While reviewing some categorical versions of the axiom of choice, it occurred to me that none of the formulations I'm aware of actually reflect how I use choice in practice: pronounce that we 'choose ...
2 votes
1 answer
308 views

Global choice and skeletons of large categories

It is stated on the nlab that the axiom of choice is equivalent to the statement that all small categories have a weak skeleton, meaning a skeletal category which is equivalent to them. Is the axiom ...
7 votes
0 answers
568 views

Automorphism groups of the complex numbers, and other fields

If one accepts the Axiom of Choice (AC), then the automorphism group of $\mathbb{C}$ is a huge and wild group, very poorly understood. But apparently if one does not accept the Axiom of Choice, then ...
4 votes
0 answers
122 views

Cofinality without choice: can this coarse definition suffer badly?

This is a rephrased version of a question previously asked at MSE without success. Working in $\mathsf{ZF}$, it is no longer possible in general to give every linear order an ordinal cofinality. For ...
15 votes
2 answers
329 views

Do we need full choice to "efficiently" use (sub)bases?

This question was previously asked and bountied at MSE without success. Suppose $(X,\tau)$ is a topological space, $B$ is a base for $\tau$, and $U\in \tau$ is an open set. Consider the following two ...
5 votes
0 answers
207 views

Which very large cardinals are preserved under Woodin's forcing for $\mathsf{AC}$?

Woodin showed that we can force $\mathsf{AC}$ if there is a proper class of supercompact cardinals while preserving supercompacts, by forcing Easton-support iteration of $\operatorname{Col}(\kappa,<...
9 votes
0 answers
1k views

What are the known implications of “There exists a Berkeley cardinal”?

Inspired by this question: What are the known implications of "There exists a Reinhardt cardinal" in the theory "ZF + j"? Definitions: $\delta$ is Berkeley iff for every $\alpha\...
4 votes
1 answer
393 views

Does $H\vDash AC$

The set $H_\kappa$ of sets hereditarily of cardinality less than $\kappa$ is defined as $H_\kappa=\{x||tc(x)|\lt\kappa\}$. What if we define the set $H=H_{Ord}$ of sets hereditarily of cardinality ...
6 votes
0 answers
237 views

Models of $\mathsf{ZF^-_2}$ over $\mathsf{ZF}$

Let $\mathsf{ZF^-_2}$ be a second-order ZF without Powerset, and the second-order Collection in place of Replacement. It is easy to see that every transitive model of $\mathsf{ZF^-_2}$ is closed under ...
11 votes
1 answer
472 views

Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$

The choice principle $\text{AC}_{\text{WO}}$ proves a large amount of cardinal arithmetic. It's well-known to imply DC, that successor cardinals are regular, and that for all $X$, there is $\lambda$ ...
10 votes
1 answer
417 views

What is first-order logic with Dedekind-finite sets of variables?

The usual set up of first-order logic is with an infinite reservoir of variables which we can use in formulas. This is one of the annoying reasons why we need to put $\aleph_0$ into the cardinal ...
42 votes
2 answers
2k views

Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?

If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps, then is there an isomorphism $V \cong W$? If we assume the axiom of choice, the answer is yes: use the fact that every ...
33 votes
3 answers
2k views

Wiki for consequences of axiom of choice?

I raised the following question as part of another MO question, but I am following the suggestion of Nate Eldredge to make it a question in its own right. For many years, there has a been a valuable ...
34 votes
1 answer
3k views

Is the theory Flow actually consistent?

Recently the paper Adonai S. Sant'Anna, Otavio Bueno, Marcio P. P. de França, Renato Brodzinski, Flow: the Axiom of Choice is independent from the Partition Principle, arXiv:2010.03664 appeared on ...
7 votes
3 answers
867 views

BCT equivalent to DC

Do you know where I can find proof of equivalence Baire Category Theorem and DC (Axiom of Dependent Choice)? It is well known fact but I can't find appropriate literature with the proof.
12 votes
0 answers
449 views

Does Hahn-Banach for $\ell^\infty$ imply the existence of a non-measurable set?

Working over ZF but without the Axiom of Choice (AC), assume that the Hahn–Banach Theorem holds for $\ell^\infty$. Does it follow that there exists a set of real numbers that is not Lebesgue ...
17 votes
1 answer
407 views

Axiom of Countable Choice and meager sets

Let us recall that the Axiom of Countable Choice (denoted by ACC) says that the countable product $\prod_{n\in\omega}X_n$ of nonempty sets $X_n$ is nonempty. It is easy to see that ACC implies that ...
10 votes
0 answers
270 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: i....
25 votes
4 answers
1k views

Is Monsky's theorem dependent on the axiom of choice?

The extension of the 2-adic valuation to the reals used in the usual proof clearly uses AC. But is this really necessary? After all, given an equidissection in $n$ triangles, it is finite, so it ...
10 votes
4 answers
958 views

On a weak choice principle

[PLEASE SEE EDITS AT BOTTOM OF QUESTION] Consider the following set-theoretic axiom: For each set $X$ there exists a set-indexed collection $\{C_i \to X\}_{i\in I_X}$ of surjections such that for ...
12 votes
1 answer
475 views

The strength of "There are no $\Pi^1_1$-pseudofinite sets"

For $\Gamma$ a set of second-order sentences in the empty language, say that a set $X$ is $\Gamma$-pseudofinite if $X$ is infinite but for every sentence $\varphi\in\Gamma$ which is satisfied in every ...
5 votes
1 answer
140 views

Maximality principle in symmetric extensions

Let $M$ be a ctm and $P\in M$ a forcing order. In regular forcing extensions, we have the following well-known Principle: $$p\Vdash_{M,P}\exists x\phi[x]\;\Longrightarrow\;\exists\sigma\in M^P\;p\...
26 votes
3 answers
2k views

Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?

The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the ...
18 votes
0 answers
359 views

Čech functions and the axiom of choice

A Čech closure function on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$,...
3 votes
1 answer
802 views

Implications of the existence of a pair of surjective functions, without Axiom of Choice

The classical Cantor-Schroder-Bernstein Theorem says that there exists a bijective function $X\leftrightarrow Y$ if and only if there exist injective functions $X\hookrightarrow Y$ and $Y\...
5 votes
0 answers
267 views

For which classes of metric spaces can we prove that quasi-isometry is an equivalence relation in ZF?

Given two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, a map $\phi \colon (M_1, d_1) \to (M_2, d_2)$ is a large-scale Lipschitz essentially surjective map if there exist constants $A \geq 1, B \geq 0$,...
5 votes
1 answer
302 views

Is Axiom of Choice equivalent to its version for families of sets, indexed by ordinals? [duplicate]

Is Axiom of Choice equivalent to the following statement? Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function $...
0 votes
0 answers
55 views

Non-uniqueness of (Galerkin) approximations and convergent subsequences without the axiom of choice?

Suppose I have an equation in some reflexive separable Banach space $X$: $$Au=f$$ for given data $f \in X^*$ and $A\colon X \to X^*$ a pseudo-monotone operator. Existence can be proved via Galerkin ...
23 votes
5 answers
6k views

Hahn-Banach without Choice

The standard proof of the Hahn-Banach theorem makes use of Zorn's lemma. I hear that, however, Hahn-Banach is strictly weaker than Choice. A quick search leads to many sources stating that Hahn-Banach ...
8 votes
0 answers
307 views

A $\mathsf{ZF}$ example of a nonreflexive group which is isomorphic to its double dual?

Given a group $G$ denote by $G^\ast=\mathrm{Hom}(G,\Bbb Z)$ its dual and by $j\colon G\to G^{\ast\ast}$ the canonical homomorphism $g\mapsto (f\mapsto f(g))$. A group is reflexive iff $j$ is an ...
10 votes
0 answers
288 views

Undetermined Banach-Mazur games: beyond DC

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; ...
4 votes
1 answer
179 views

Distributive lattices and axiom of choice

What form of the axiom of choice is equivalent (in ZF) to the statement that every distributive lattice is isomorphic to a lattice of sets?
5 votes
1 answer
513 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 ...
19 votes
4 answers
3k views

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 tells ...
1 vote
0 answers
105 views

Can small class choice be weaker than global choice and stronger than set choice + collection?

In this posting what was termed as "Proper Class Choice" principle turned to be equivalent to Global Choice over the base theory of "MK-Foundation -Limitation of size + Set Replacement*&...
6 votes
0 answers
382 views

General theory of the reals in Solovay-like models

Solovay's model is a famous model of $\sf ZF$ where we start in $L$ with $\kappa$ inaccessible, and we collapse all the ordinals below $\kappa$ to be countable, without collapsing $\kappa$ itself, and ...
-1 votes
1 answer
239 views

Is Proper Class Choice equivalent to Global Choice?

Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom: Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \...
10 votes
1 answer
496 views

Models of ZF intermediate between a model of ZFC and a generic extension

Let $M$ be a countable model of $ZFC$ and $M[G]$ be a (set) generic extension of $M$. Suppose $N$ is a countable model of $ZF$ with $$M\subseteq N \subseteq M[G]$$ and that $N=M(x)$ for some $x\in ...
9 votes
0 answers
355 views

A $\mathsf{ZF}$ example of two Baire spaces whose product is not Baire?

Motivated by this question I'm looking for a pair of Baire topological spaces whose product is not Baire and whose construction does not need the axiom of choice. The example of such spaces I'm ...
19 votes
4 answers
1k 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 ...
2 votes
1 answer
244 views

Can the axiom of choice or its weaker versions be (dis)proved using reflection principles?

In On the Question of Absolute Undecidability, Peter Koellner investigates whether it is possible to prove or disprove $V = L$ using (EDIT: both first and second-order) reflection principles, ie. ...
8 votes
1 answer
222 views

Aronszajn Trees when AC fails

This question may be easy and indicative of my ignorance about the failure of the axiom of choice. If so, I apologize. Below assume $\mathsf{DC}$ but not $\mathsf{AC}$. Suppose we have a partial order ...
6 votes
1 answer
811 views

Very large axiom of choice

let me say that I am not a set theorist, but I have to settle up some things in category theory and I need your help. What I'd like to do is, in some way, use axiom of choice for proper classes. I ...
1 vote
1 answer
259 views

Does choice always hold in a model of ZF with point-wise parameter-free definable sets?

If one add to ZF the rule that all sets are parameter free definable. Would that prove the axiom of choice? More specifically. IF we add the following omega rule to inference rules of the language of ...
14 votes
0 answers
306 views

Complement-like operator and the axiom of choice

We say that an operator $^*$ on ${\cal P}(A)$ is $\star$-complement if $^*$ is not the complement operator and for all $X⊆A$ we have: $X^*∪X=A$ $X^{**}=X$ We say that $^*$ is $\star$-strong ...
2 votes
1 answer
858 views

Are closed convex subsets of a Banach space weakly closed without the axiom of choice?

It is a well-known fact that closed convex sets in Banach spaces are weakly closed. The common proof is based on the Hahn-Banach theorem that uses the axiom of choice. Is there any proof of this fact ...
5 votes
0 answers
166 views

How much choice is required for a countably-infinite index subgroup of the real additive group?

The existence of such subgroups implies the existence of a non-measurable set; simply intersect each of the cosets with $[0,1]$. The results will all have equal outer measure, but their union will be ...
16 votes
1 answer
705 views

The axiom $I_0$ in the absence of $AC$

It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$ then $\lambda$ has countable cofinality (and in ...
1 vote
0 answers
206 views

Uncountable chain of nested sets without choice

Let $\kappa$ be an uncountable cardinal. Given a set of S of cardinality $\kappa$, I want to construct a chain {$S_\lambda : \lambda \in \kappa$ } such that: 1) Each $S_\lambda$ is a proper subset of ...
8 votes
0 answers
362 views

Can the set of endomorphisms of $(\mathbb R,+)$ have cardinality strictly between $\frak c$ and $\frak{c^c}$?

Let $\frak c$ be the cardinality of the reals. I know that in ZF the set of endomorphisms of $(\mathbb R,+)$ can have at least two different cardinalites: If we allow the axiom of choice, you can ...

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