# Tagged Questions

The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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### Uncountable dense sets of reals without the axiom of choice

In the absence of AC, can there be an uncountable dense set $S\subset\mathbb R$ such that $S\cap(-\infty,a)$ is countable for each real number $a$? (Of course, since $S$ is a countable union of ...
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### Choice function without AC in special case [duplicate]

I read the Jech, Set theory, and saw following proposition. (☆) If S is a finite family of nonempty sets, existence of choice function of S can be proved without axiom of choice. I tried to prove ...
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### If every partitioning of $X$ has a choice function, is $X$ necessarily well-orderable?

Working over the ZF axioms, it's clear that if $X$ is a well-orderable set, then every partitioning of $X$ has a choice function, by choosing the minimum of each cell. Question. Does the converse ...
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### Equivalence of Taylor series and its corresponding function and Axiom for infinite summation

Given a function $f(x)$ with a taylor series expansion, is it valid to say that $$f(x)=\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(a)(x-a)^n$$ for all values of x irrespective of whether the taylor series ...
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### Is it really necessary to choose a basis to prove that every element of a comodule is contained in a finite-dimensional subcomodule?

Let $C$ be a coalgebra over a field and $M$ be a $C$-comodule. Then it's well-known that every element of $M$ is contained in a finite-dimensional subcomodule $M' \subset M$. This is for example an ...
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### Impossibility of constructing a continuum-size linearly independent set in $\Bbb R$ [duplicate]

This is a response to the following exchange at Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional? [Bill constructs a $\aleph_0$ ...
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### A set that satisfies the hypothesis of Zorn's Lemma

A set $x$ satisfies the hypothesis of Zorn's Lemma. Let $k \in x$. $\textbf{Prove:}$ There is a $z \in x$ such that $z$ is $S$-maximal in $x$ and $z=k \vee kSz$ $\textbf{Attempt:}$ ...
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### When can “$j: V \rightarrow M$ is an elementary embedding” be defined in ZF?

This regards elementary embeddings of inner models of set theory. It seems that it is in general "stated" via an axiom schema each member of which states that the class function is elementary with ...
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### Bolzano-Weierstrass Theorem Proof Abbott (Ax. of Choice)

In Abbott's Proof of the Bolzano-Weierstrass Theorem, does Abbott use any form of the Axiom of Choice ? I think he is since he chooses an $a_{n_k} \in I_k$ where there are multiple such $a_{n_k}$.
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### Axiom of choice is equivalent to every relation includes a function with the same domain

The axiom of choice asserts that for any set $X$ there exists a function $f : (2^X − \{\emptyset\}) \rightarrow X$ such that for any nonempty $A \subseteq X$, $f(A) \in A$. Show that this is ...
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### Can a basis for $\mathbb{R}$ be Borel?

Work in ZF (so no choice). Then it is consistent that there is no (Hamel) basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space. My question is about models where $\mathbb{R}$ does have a basis, but ...
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### Do the relative consistency results involving the axiom of choice use completeness of FOL?

The proof of the Completeness Theorem for first order logic uses the axiom choice and is a central result in model theory. Chang and Keisler claim that model theory has important applications to set ...
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### Spanning the reals with a small set - choicelessly

Working in ZF (so, no choice): is it possible that there is a set of reals $X$ such that $\vert X\vert<\mathbb{R}$, but $X$ generates $\mathbb{R}$ as a subgroup under addition? This seems ...
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### Surjective iff right invertible. Bijective iff invertible [duplicate]

Let $f:X \to Y$ be a mapping. Its known that choice implies $f$ is surjective iff $f$ is right invertible. Are the following statements true? Surjective iff right invertible is equivalent to choice. ...
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### Does the axiom of choice have any use for finite sets?

It is well known that certain properties of infinite sets can only be shown using (some form of) the axiom of choice. I'm reading some introductory lectures about ZFC and I was wondering if there are ...
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### Axiom of choice - Equivalence relation - Representatives

Let $X$ be a set and $\sim$ an equivalence relation on $X$. In many proofs, a set of representatives of equivalence classes of $X$ is used (e.g. coset or orbit representatives in groups, ...
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### Every set of pairwise disjoint sets has a choice function implies AC

Suppose that for every collection $A$ of non empty pairwise disjoint sets has a choice function. I need to prove that this implies the axiom of choice. Let $S$ be a collection of sets. For every $B$ ...
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### Why is the statement “all vector space have a basis” is equivalent to the axiom of choice? [duplicate]

I'm reading a section in an abstract algebra book, where it reviews vector spaces and suddenly comments that "all vector space have a basis" is equivalent to the axiom of choice...I haven't studied ...
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### Ordinal enumeration in ordered Mostowski model - does it not need the global choice?

In Jech's axiom of choice he proves following lemma (lemma 4.5(b) in his book): There is a injective mapping from $M$ to $\mathrm{Ord}\times \operatorname{fin}(A)$, where $A$ is a set of atoms and ...
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### Show that the sets of isolated point of $E$ is a countable set - Axiome of choice

Let $E \subset \mathbb{C}$. Show that the sets of isolated point of $E$ is a countable set. That question is related to this question. However, my question somewhat different. Define $Acc(E)^c$ ...
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### Does the following condition implies full outer measure?

Let $X \subseteq 2^{\omega}$ be a set of positive Lebesgue measure. Suppose that for every $\eta, \nu \in 2^{<\omega}$ of the same length, the measure of $X$ above $\eta$ is the same as the measure ...
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### What does Axiom of Choice mean [duplicate]

So this a fundamental assumption in mathematics. Can someone explain informally what it actually is please. My guess is that its when we say in proofs that "Let $x \in X$". But I am not sure.
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### Prob. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian product

The Axiom of Choice is as follows: Given a collection $\mathcal{A}$ of disjoint non-empty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathcal{A}$; that ...
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### Well-foundedness of cardinals and the axiom of choice

Without axiom of choice it is not generally true that the class of all cardinal (in this question we consider Scott cardinal rather than cardinals as ordinals) is not well-founded under the ordinary ...
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### Algebraic closure with no nontrival automorphism

In Milne's notes on Galois theory, Chapter 7, p.91 he remarked that it is consistent without the axiom of choice that there exists an algebraic closure $L$ of $\mathbb{Q}$ with no nontrivial ...
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### Different Zorn's lemma statements

Given a chain-complete poset $P$, every $x\in P$ lies below some maximal element. Every inductive poset has enough maximal elements a maximal element. Chain-complete means every chain has a least ...
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### Can a large $V_\alpha$ satisfy Comparability but not be well-orderable?

Say that a set satisfies Comparability if any two of its subsets are comparable: one is injectable into the other. Are there models of ZF containing ranks $V_\alpha$ which satisfy Comparability but ...