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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Class-theoretic division paradox

The Division Paradox is the fact that there are models of ${\sf ZF \neg C}$ in which a set can be partitioned into a set that is bigger than it — equivalently, in which there are sets $X$ and $Y$ such ...
user171348's user avatar
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Basic cardinal arithmetic without choice

Do we know everything about addition and multiplication of cardinalities in choiceless set theory? For example, let $M$ be a model of $\textsf{ZF}+\textsf{AD}+V=L(\mathbb{R})$, consider the sets $\...
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14 votes
1 answer
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Stone-Weierstrass Theorem without AC

To what extent does the usual Stone-Weierstrass Theorem depend on some form of the Axiom of Choice? There seems to be a lot of literature on constructive versions in toposes, but I have been unable ...
Bruce Blackadar's user avatar
2 votes
0 answers
96 views

Elementary abelian 2-subgroups of $\mathrm{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ (with and without choice)

Consider the absolute Galois group $G_{\mathbb{Q}} := \mathrm{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$. As I understand it, the only torsion elements have order $2$ (by Artin-Schreier), and they are ...
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What is the strongest form of the Axiom of Choice available in $\mathsf{Z}_{2}$?

$\mathsf{Z}_{2}$ denotes second-order arithmetic. Some forms of AC are expressible in $\mathsf{Z}_{2}$; for example the $\mathsf{\Sigma}_{1}^{1}$ axiom of choice is part of the theory $\mathsf{ATR}_{0}...
Victor's user avatar
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5 votes
1 answer
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Large cardinals in ZF + DC + AD

The Axiom of Dependent Choice (DC) is often considered to be an "intuitive and non-controversial" version of choice used in the proofs of many theorems in Analysis. Similarly, the Axiom of ...
user42761's user avatar
6 votes
1 answer
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Automorphisms of algebraically closed fields without the Axiom of Choice

In the paper Algebraische Konsequenzen des Determiniertheits-Axioms (U. Felgner – K. Schulz, Arch. Math. (Basel) 42 (1984), pp. 557–563), the authors show that in models of Zermelo-Fraenkel set theory ...
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Weak trichotomy principle in the absence of choice

It is well known that the trichotomy property of cardinals ($∀κ,λ\in\operatorname{Card}(κ<λ∨κ=λ∨κ>λ)$) is equivalent to the axiom of choice. D. Feldman and M. Orhon had defined in [1] a ...
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-2 votes
1 answer
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Does cardinal definable choice imply AC?

Recall the definition of cardinal definable sets, to re-iterate: $Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda_1,.., \...
Zuhair Al-Johar's user avatar
4 votes
1 answer
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Exactly how much (and how little) can partial ordered sets (classes) embed to the cardinalities

In the paper "Convex Sets of Cardinals", Truss mentioned a result of Jech: If $M$ is a countable transitive model of ZFC, and $(P,<)∈M$ is a poset, then there exists a Cohen extension of ...
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Does n-well ordered choice schema imply the axiom of choice?

Define: $\operatorname {wo}^n(x) \iff \forall y (y \in^n x \to \operatorname {wo} (y))$ Where $\operatorname {wo}(y)$ refers to $y$ being well orderable. Where $y \in^0 x \iff y=x \\ y \in^{n+1} x \...
Zuhair Al-Johar's user avatar
4 votes
1 answer
260 views

Automorphisms of vector spaces and the complex numbers without choice

In Zermelo-Fraenkel set theory without the Axiom of Choice (AC), it is consistent to say that there are models in which: there are vector spaces without a basis; the field of complex numbers $\mathbb{...
THC's user avatar
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Is this version of Nested Selection equivalent to AC?

This is an endeavor to salvage "Nested Selection" principle presented in a prior posting. Define $ \begin{align} Y \text { is } \Phi \text{-image of } X \iff &\forall a \in X \exists b \...
Zuhair Al-Johar's user avatar
-4 votes
1 answer
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Is Nested Selection equivalent to AC?

Nested Selection: For every infinite set $G$ of pairwise disjoint infinite sets such that any two distinct elements $x,y$ of $G$ either "$y$ is a set of proper supersets of elements of $x$ and ...
Zuhair Al-Johar's user avatar
2 votes
1 answer
56 views

Infinite decreasing sequence for class relation without minimal elements

Let us assume $<$ is some class relation without minimal elements, meaning $\forall a, \exists b, b< a$. This means that for every $n\in\omega$, one can build a decreasing function $f$ with ...
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Does the Weak Power Hypothesis imply the Boolean Prime Ideal Theorem?

If there is a bijection $\varphi:x\to y$ between two sets $x$ and $y$, we use the notation $x\simeq y$. The Weak Power Hypothesis is the following statement: (WPH) For all sets $x, y$, whenever ${\...
Dominic van der Zypen's user avatar
6 votes
2 answers
409 views

Can Deep Choice entail Axiom of Choice?

Deep Choice: $\forall X \ [\forall a,b \in X \, ( a \neq \emptyset \land (a \neq b \to a \cap b = \emptyset)) \to \\ \exists Y \exists f \,(f: X \rightarrowtail Y \land \forall x \in X \,( f(x) \in \...
Zuhair Al-Johar's user avatar
1 vote
0 answers
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How much choice we can get from this partition principle?

For every set $X$ there cannot be a partition on $X$ of a larger size than the set $\iota``X$ of all singleton subsets of $X$. Formally: $$\begin{align} \forall X \forall P: P \text { is a partition ...
Zuhair Al-Johar's user avatar
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Does parallelism of cardinal comparison between sets and their power sets, enact a form of choice? [duplicate]

Let $ * $ range over cardinal relations $ \{<,<>\}$; if we add the following axiom to $\sf ZF$, would that prove a known form of choice? Parallelism: $ |x| * |y| \leftrightarrow |\mathcal P(...
Zuhair Al-Johar's user avatar
2 votes
0 answers
67 views

Is "strongly unbounded logics are unbounded" equivalent to "no descending sequence of cardinals"?

This question is motivated by Vaananen's paper Generalized quantifiers in models of set theory. Say that a (set-sized, regular) logic $\mathcal{L}$ is unbounded if there are $\mathcal{L}$-sentences $\...
Noah Schweber's user avatar
4 votes
0 answers
208 views

Can we have full choice prior to Reinhardt cardinals?

Working in $\sf ZF + Reinhardt \ cardinal$, can we have full choice over all stages $V_{\alpha < \kappa}$ where $\kappa$ is the Reinhardt cardinal, i.e., the critical point of the elementary ...
Zuhair Al-Johar's user avatar
1 vote
0 answers
53 views

Can this method let choiceless large cardinals be smaller than cardinals compatible with choice?

Recall question "Can we have this sequence where choice fails and returns?" Can that theory be extended with requiring the $\mathcal V_n$'s to fulfill a choiceless large cardinal extension ...
Zuhair Al-Johar's user avatar
4 votes
1 answer
203 views

Can we have this sequence where choice fails and returns?

Can we have a sequence of transitive sets $\langle\mathcal V_0, \mathcal V_1, \mathcal V_2,...\rangle$, all modeling $\sf ZF$, such that $\mathcal P(V_n) \subset \mathcal V_{n+1}$, and the cardinality ...
Zuhair Al-Johar's user avatar
22 votes
8 answers
2k views

Simpler proofs using the axiom of choice [duplicate]

I am looking for examples of results which may be proven without resorting to the axiom of choice/Zorn lemma/transfinite induction but whose proof is quite simplified by the use of the axiom. For ...
4 votes
1 answer
649 views

About the relationship between the generalized continuum hypothesis and the axiom of choice

I was trying to get a short, intuitive proof of Sierpinski’s theorem (gch implies axiom of choice) and I could but only by using the following gch2 for the generalized continuum hypothesis gch. gch: ...
M. Solomon's user avatar
8 votes
1 answer
927 views

Is there a form of choice that can elude Kunen's inconsistency theorem?

When it is said that Kunen inconsistency theorem proves that given $\sf ZFC$ no elementary embedding can exist from the universe to itself. Most references quote full choice in stating that result, ...
Zuhair Al-Johar's user avatar
12 votes
2 answers
743 views

Ultrafilter lemma for arbitrary lattice

Can someone kindly confirm whether the ultrafilter lemma for arbitrary (i.e., not necessarily Boolean) bounded lattices is equivalent to Zorn's lemma? To be precise, if $\mathbf{L} = (L, \leq, \land, \...
Menander I's user avatar
11 votes
0 answers
340 views

On the division paradox

This question is partly motivated by Timothy Chow's recent question on the division paradox. Say that a set $X$ admits a paradoxical partition if and only if there is an equivalence relation $\sim$ on ...
Andrés E. Caicedo's user avatar
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0 answers
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Irreducible subcontinuum without Zorn's lemma

In continuum theory we frequently use the fact that two points in a continuum are contained in an irreducible subcontinuum. A continuum $X$ is a compact connected metric space. A subcontinuum $K\...
D.S. Lipham's user avatar
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3 votes
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Weak Power Hypothesis with injections instead of bijections

Let $x,y$ be sets. We use the following notation: $x\simeq y$ means that there is a bijection $\varphi:x\to y$, and $x\leq y$ means that there is an injection $\iota:x\to y$. The Weak Power ...
Dominic van der Zypen's user avatar
7 votes
1 answer
388 views

Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets

There is an old result due to Mycielski and Sierpiński, and popularized in a Monthly article by Taylor and Wagon (A Paradox Arising from the Elimination of a Paradox; see also this MO answer), that ...
Timothy Chow's user avatar
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46 votes
7 answers
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Zorn's lemma: old friend or historical relic?

It is often said that instead of proving a great theorem a mathematician's fondest dream is to prove a great lemma. Something like Kőnig's tree lemma, or Yoneda's lemma, or really anything from this ...
Pace Nielsen's user avatar
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8 votes
1 answer
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Compatibility of $\mathsf{SVC}$ and Reinhardtness

Can we prove the consistency of $\mathsf{ZF+SVC}$ + "There is a Reinhardt cardinal?" (Preferably from the consistency of $\mathsf{ZF}$ with a Reinhardt cardinal, but using a stronger ...
Hanul Jeon's user avatar
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8 votes
1 answer
375 views

Automorphisms of projective spaces, and the Axiom of Choice

It is known that upon not accepting the Axiom of Choice (AC), there exist models of ZF in which there are projective spaces (over a division ring) with a trivial automorphism group. (This is a truly ...
THC's user avatar
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8 votes
1 answer
193 views

Some relevant questions about the consistency strength of singularity of $\omega_1$ and $\omega_2$

The following question was asked years ago on MSE, but let me recap it: Question: Is there anything currently known about the exact consistency strength of "$\mathsf{ZF}$ + both $\omega_1$ and $\...
Hanul Jeon's user avatar
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12 votes
1 answer
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Is there a more modern account of the main results of "Adding Dependent Choice" by D. Pincus?

I would like to read Pincus' article Adding dependent choice, where he proves, among other things, the consistency of the theory $\mathsf{ZF+DC+O+\neg AC}$, where $\mathsf{DC}$ stands for Dependent ...
Lorenzo's user avatar
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13 votes
0 answers
253 views

Converse of Knaster-Tarski's theorem as choice principle

Knaster-Tarski's theorem States that if $(A,\le)$ is a complete lattice, then every monotone function $f :A \to A$ has a fixed point. The proof is carried out in $\mathsf{ZF}$. By $\mathsf{KTC}$ we ...
Average-user's user avatar
7 votes
1 answer
289 views

How hard is it to get "absolutely" no amorphous sets?

A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets&...
Noah Schweber's user avatar
13 votes
0 answers
288 views

$\mathsf{AC}_\mathsf{WO}+\mathsf{AC}^\mathsf{WO}\Rightarrow \mathsf{AC}$?

Let $\mathsf{AC}_\mathsf{WO}$: Every well-orderable family of non-empty sets has a choice function. $\mathsf{AC}^\mathsf{WO}$: Every family of non-empty well-orderable sets has a choice function. My ...
Lorenzo's user avatar
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8 votes
0 answers
292 views

Why is the double negation of the axiom of choice rarely considered?

In constructive/intuitionistic mathematics, it is common to reject the axiom of choice, because it is highly nonconstructive and implies the law of the excluded middle by Diaconescu's theorem/Bishop's ...
saolof's user avatar
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4 votes
0 answers
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$2^{|V|}$ class cardinalities without global choice

Is it consistent with Morse-Kelley set theory without global choice (but with choice for sets) that there are $2^{|V|}$ proper classes of different cardinalities? Alternative question: Is it ...
Dmytro Taranovsky's user avatar
1 vote
0 answers
148 views

Can we have a spectrum of intermediate choice properties between set choice and global choice?

Global choice is equivalent to saying $|V|=ON$, while ordinary choice is equivalent to saying that $V= H_{< ON}$. So the relative cardinalities of $V$ and $ON$ seems to affect the degree of choice ...
Zuhair Al-Johar's user avatar
7 votes
0 answers
296 views

Is there a "nice" inner model for $\mathsf{ZF}$ + a Dedekind-finite infinite set of reals?

Below, given a formula $\varphi$ which $\mathsf{ZF}$ proves defines a set of reals and an inner model $W$, I'll write "$\varphi^W$" and "$L(\varphi^W)$" for "$\{x:W\models\...
Noah Schweber's user avatar
12 votes
4 answers
2k views

What can be preserved in mathematics if all constructions are carried out in ZF?

This is inspired by this discussion. I see that the debates about the necessity of the axiom of choice in this or that statement are still ongoing. In this regard, I became interested in whether there ...
Sergei Akbarov's user avatar
3 votes
0 answers
96 views

In characterizing finiteness via $\mathsf{AC}_\omega$, do we need to use large sets?

The usual proof that "Dedekind finite = finite" from $\mathsf{ZF+AC_\omega}$ goes by, given a Dedekind-infinite set $X$, applying countable choice to the sequence $(X^n)_{n\in\omega}$. It ...
Noah Schweber's user avatar
21 votes
1 answer
988 views

A Krull-like Theorem and its possible equivalence to AC

A well known equivalent of the Axiom of Choice is Krull's Maximal Ideal Theorem (1929): if $I$ is a proper ideal of a ring $R$ (with unity), then $R$ has a maximal ideal containing $I$. The proof is ...
Michael Kinyon's user avatar
9 votes
0 answers
197 views

Is any choice axiom other than WISC inherited by Grothendieck topoi?

It is well known that even if one works with say ZFC as a base theory, Grothendieck topoi do not in general satisfy even fairly weak axioms like countable choice or small violations of choice and one ...
saolof's user avatar
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4 votes
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Is this "finite-ish combinatorics" reflection principle consistent?

This question is an attempt to chisel away at this earlier question of mine, which in retrospect may be rather intractable. Throughout, we work in $\mathsf{ZF}$. Briefly (see the linked question for ...
Noah Schweber's user avatar
4 votes
0 answers
133 views

Consistency of a strange (choice-wise) set of reals, pt. 2

This is a follow-up on this question. Consider a set $X\subseteq \mathbb{R}$ such that $X$ is not separable wrt its subspace topology Every countable family of non-empty pairwise disjoint subsets of $...
Lorenzo's user avatar
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7 votes
2 answers
672 views

Consistency of a strange (choice-wise) set of reals

Consider a set $X\subseteq \mathbb{R}$ such that $X$ is not separable wrt its subspace topology For all $r\in\mathbb{R}$ there exists a sequence $(x_n)_{n\in\omega} \subset X$ converging to $r$ In a ...
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