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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17
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237 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) $\varphi(\emptyset)=\emptyset$,...
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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\...
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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 $A \geq 1, B \geq 0$,...
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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 $...
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Does the “three-set-lemma” imply the Axiom of Choice?

Consider the following curious statement: $(S)$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \...
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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 ...
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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 ...
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124 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?
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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; ...
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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*". However if ...
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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 ...
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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 \subset R \ (...
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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 ...
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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 ...
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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. ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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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366 views

Failure of SVC in Grothendieck toposes

The axiom SVC (for "small violations of choice") asserts that there is a set $S$ such that for every set $X$ there is a choice set $A$ such that $X$ is a subquotient of (i.e. admits a surjection from ...
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Existence of a certain set of 0/1-sequences without the Axiom of Choice

Is there a set $\mathcal X\subset\{0,1\}^{\Bbb N}$ of 0/1-sequences, so that For any two 0/1-sequences $x,y\in\{0,1\}^{\Bbb N}$ for which there is an $N\in\Bbb N$ with $$x_i=y_i,\;\;\text{for all $i&...
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Polish transversals

A subset of $X$ an indecomposable continuum $Y$ is called a composant transversal if $X$ has exactly one point from each composant of $Y$. So a continuum has a composant transversal precisely when ...
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264 views

Exterior powers and choice

Under the assumption that any vector space has a basis (so under the assumption of the axiom of choice), we can prove the following algebraic statements : 1) If $\varphi:V\to W$ is an injective ...
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Relationship between AC, WO, and Zorn's lemma in ZF-Powerset

In regular ZF, AC, WO, and Zorn's Lemma are equivalent, but every proof I know (of the implication AC -> WO and AC -> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is ...
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Injection into a proper class and choice without regularity

In $\sf ZF$, we have that the axiom of choice is equivalent to: For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$ and For all sets $X$, and for all proper classes $Y$, $Y$ ...
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Measure of rational hyperplanes of $\mathbb{R}$

Let's view $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and pick some basis $(v_\alpha)_{0 \leq \alpha < \mathfrak{c}}$ of it. We can then consider the subspace $L$ spanned by $(v_\alpha)_{0 &...
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472 views

Cardinal characteristics of amorphous sets

In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this ...
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Amorphous proper classes in MK

Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be ...
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How many algebraic closures can a field have?

Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is ...
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Is Global Choice conservative over Zermelo with Choice?

To be explicit, by Zermelo set theory with Choice, ZC, I mean the theory with the same language and axioms as ZFC except not Foundation (also called Regularity) and with the axiom scheme of Separation ...
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Axiom of choice and algebraic tensor product

The first part of the question was asked on Math-stackexchange. Let $V$, and $W$ be vector spaces. By the universal property of the tensor product, there is a canonical map from $V^*\otimes W^*$ ...
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Is this lemma equivalent to the axiom of choice?

Given any pre-ordering $\preceq$ of an arbitrary set $X$ is the following lemma: $$\text{There exists an inclusion minimal set }S\text{ satisfying }\{a\preceq b:b\in S\}=X\\\iff \text{ Every chain in ...
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Maximizing “happy” vertices in splitting an infinite graph

This question is motivated by a real life task (which is briefly described after the question.) Let $G=(V,E)$ be an infinite graph. For $v\in V$ let $N(v) = \{x\in V: \{v,x\}\in E\}$. If $S\subseteq ...
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Surreal numbers and the Axiom of Choice

In ZFC and its conservative extension NBG, it can be shown that every ordered field embeds into the surreal numbers. How much choice is needed to prove this? Without choice, what is a simple example ...
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Stronger negation of AC given by rejecting “infinite hat” puzzles

Some of the strangest implications of AC are the "infinite hat" puzzles, which are on Wikipedia, and have been talked about on MO several times, including some variants. There are different ways to ...
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Axiom of Choice versus V=L in opposition to large cardinals

Consider the following two observations: The axiom $V=L$ is incompatible with large cardinal axioms that are somehow "too large", like measurable cardinals. The axiom of Choice is incompatible with ...
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Compactification of Tychonoff spaces without full axiom of choice

If $X$ is a Tychonoff space, then using the Tychonoff theorem and thus the full axiom of choice, it follows that $X$ admits a Hausdorff compactification. My question is : what remains true if we do ...
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Relation between the Axiom of Choice and a the existence of a hyperplane not containing a vector

In a lot of problems in linear algebra one uses the existence, for each $E$ vector space over a field $k$, and each $x\in E$, of a Hyperplane $H$ such that $E=k\cdot x \oplus H$ (Let us denote $\...
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Cardinality vs. isomorphism type of vector spaces without choice

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem: If $V$ is an infinite vector space over a field $F$, and $...
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Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets?

Is the following consistent with $\text{ZF}$? There exists a set $S=\{x_1,x_2,x_3,...\}$ such that: $|x_{i+1}| < |x_i|$ $\forall m,n \in S (|P(m)|=|P(n)|)$ Where cardinality $``||"$ is ...
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Are there known examples of sets whose power set is equal in size to power set of larger sets only in absence of choice?

The question of existence of sets $x,y$ such that $$|x|<|y| \wedge |P(x)|=|P(y)|$$ is known to be independent of $\text{ZFC}$! But are there known examples of sets fulfilling the above condition ...
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427 views

Do choice principles in all generic extensions imply AC in $V$?

It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a ...
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The Tall Tale of Terminating Transfinite Towers

The transfinite tower of iterative automorphisms of a group $G$ is simply definied to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct ...
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The partial preorder on $\mathbb N$ generated by the finite axioms of choice

Let $\mathsf C_n$ denotes the statement: for any family $\mathcal F$ of $n$-element sets there exists a choice function (i.e., a function $f:\mathcal F\to\bigcup\mathcal F$ such that $f(F)\in F$ for ...
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The Rise and Fall of Dictators & How it Depends on Our Choice

This question is loosely inspired by the following paper of Shelah on Arrow property in which he answered a question of Gil Kalai affirmatively. Shelah, Saharon, On the Arrow property. Adv. in Appl. ...
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324 views

Countable (?) dependent choice

In some circumstances I've been using a form of choice over the first uncountable ordinal knowing a priori that only a countable number of choices were going to be made (without any a priori upper ...

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