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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Actual infinitesimals for solving Vitali paradox
Has anyone tried to use actual infinitesimals to solve paradoxes about non-measurability? In Vitali paradox, for example, they divide a set with measure 1 into $\infty$ subsets of zero measure and ...
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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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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 ...
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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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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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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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Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?
A result commonly, and probably erroneously, attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on ...
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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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