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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Hereditarily countable sets in Antifounded ZF
A set $x$ is hereditarily countable when every membership-descendant of $x$ (including $x$ itself) is countable.
In this paper, Jech proved in ZF that the class of all hereditarily countable sets is a ...
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Proof/Reference to a claim about AC and definable real numbers
I’ve read somewhere on this site (I believe from a JDH comment) that an argument in favor of AC (I believe from Asaf Karagila) is that without AC, there exists a real number which is not definable ...
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Infinitary logics and the axiom of choice
Suppose we want to enhance ZF by allowing for infinitary formulas instead of just first-order ones in our axiom schema of separation and/or replacement. It seems that we don't need much power in our ...
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Everywhere-defined unbounded operators between Banach spaces
In this post, it is said that there are no constructive examples of everywhere-defined unbounded operators between Banach spaces; every example furnished must use the axiom of choice. This seems like ...
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Is discriminative choice provable in ZFC?
Let $\phi$ be a formula defining an equivalence relation.
Definitions:
The $\phi$-cardinality of a set $X$ be the cardinality of $X/\phi$. That is, the cardinality of the set of all equivalence ...
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Complexity of definable global choice functions
It is well-known that $L$ has a $\Sigma_{1}$-definable global choice function; it is also known that there are other transitive class models of ZFC with this property.
I wonder about the complexity ...
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Is a weak version of the three sets Lemma provable in ZF?
The Three Sets Lemma is the following Lemma:
Lemma: Let $f(x)$ be a function from $X$ to $X$ where $f(x)$ has no fixed points. Then there exists a partition of $X$ into three disjoint sets $X_1$, $X_2$...
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Must strange sequences wear Russellian socks?
This is an attempt to make more precise a vague guess at the end of this answer of mine. We work in $\mathsf{ZF}$ throughout.
Say that a sequence $\mathcal{A}=(A_i)_{i\in\omega}$ of disjoint sets is ...
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Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?
Does $\sf ZFA + WOIPS$ prove $\sf AC$?
Where $\sf WOIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.
In $\sf ZF$, I ...
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Turning linear ordering into well-ordering
Work in ZF + DC. Assume that there is an uncountable $A \subseteq \mathbb{R}$ and a linear order $\prec$ on $A$ such that for every $x \in A$, $\{y \in A: y \prec x\}$ is countable. Must there exist ...
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The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)
A set $X$ is Dedekind-infinite if there is an injective map $f: X\to X$ that is not surjective.
A set $X$ is dually Dedekind-infinite if there is a surjective map $f: X\to X$ that is not injective.
In ...
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A cardinal inequality for finiteness
Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality:
$$2^{2^{2^{2^{\mathfrak{n}}}}}...
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Partition of unity without AC
Several existence theorems for partition of unity are known. For example (source),
Proposition 3.1. If $(X,\tau)$ is a paracompact topological space,
then for every open cover $\{U_i \subset X\}_{i \...
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Does $\mathsf{SVC}^\ast$ exist?
$\mathsf{SVC}(S)$ is the assertion that for all sets $X$ there is an ordinal $\eta$ and a surjection $f\colon\eta\times S\to X$. I would like to denote by $\mathsf{SVC}^\ast(S)$ the same assertion but ...
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Proof of the axiom of choice for finite sets in ZF [closed]
Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$ satisfying $...
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