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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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 ...
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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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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,.., \...
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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
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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{...
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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
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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
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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 ${\...
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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
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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 ...
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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
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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 $\...
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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
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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
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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
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8 answers
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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 ...
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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: ...
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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
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2 answers
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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
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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 ...
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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\...
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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
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1 answer
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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 ...
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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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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 ...
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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 ...
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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 $\...
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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 ...
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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
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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&...
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$\mathsf{AC}_\mathsf{WO}+\mathsf{AC}^\mathsf{WO}\Rightarrow \mathsf{AC}$?

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 ...
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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 ...
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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
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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
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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
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4 answers
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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
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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
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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 ...
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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 ...
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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
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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 $...
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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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Ramsey's infinite principle and the axiom of choice

Frank Plumpton Ramsey, best known for giving his name to Ramsey Theory, presented the following theorem in On a Problem of Formal Logic, that was submitted in 1928 and published posthumously. Let $\...
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Reference for countable and uncountable algebraic closures of $\mathbb{Q}$ in ZF

The following facts seem to be part of the folklore (where $\mathsf{ZF}$ means Zermelo-Fraenkel set theory with no axiom of choice): it is consistent with $\mathsf{ZF}$ that there exists an ...
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Does Tarski's squaring theorem imply Axiom of Choice in NFU?

I'm trying to see which results from mainstream set theory (ZF) about Axiom of Choice can be proved in New Foundations with Urelements (U is added simply because ...
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How much "finitary combinatorics" can be emulated by an infinite Dedekind-finite set?

Throughout, we work in $\mathsf{ZF}$. Let $[n]=\{1,...,n\}$. Given a set $X$ and a first-order sentence $\varphi$, let $M_X(\varphi)$ be the set of isomorphism types of models of $\varphi$ with ...
Noah Schweber's user avatar
5 votes
2 answers
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Absolute Galois group, number theory and the Axiom of Choice

Richard Taylor once explained his research (in number theory) in a very simple way: understanding the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. It is known that in ...
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