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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Conceptual problem in Noetherian rings and Zorn Lemma

I've got a problem, and it's that I can't see the difference of one of the definitions of Noetherian ring, and supposing the zorn lemma for that ring. I mean, if I use the definition of the maximal ...
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Closure of a set in a “Topology of finite complement”

Well, I was reading this article by Kelley and when reached the point where he say that $X_a$ is closed in $Y_a$ I had to stop, probably mine is just a stupid misunderstand but can't figure out how to ...
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Did I use axiom of choice in my proof?

I have two different affine open covers for a scheme $X$, say $X = \cup_{i \in I} U_i$ and $X = \cup_{j \in J} V_j$. For each $p \in X$, we know there exist some $i(p)$ and $j(p)$ such that $p \in ...
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Every epimorphism in Sets is split: why is it equivalent to axiom of choice?

Suppose that $f: A \rightarrow B$ is epic in Sets. One can construct a section $s: B \rightarrow A$ of $f$ as follow: Let us define an equivalence relation $R$ on $A$ as follow: $aRa'$ iff $a, a' \in ...
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Forcing reference

Who first proved that, over ZF, the statement (1) The reals are well-orderable is strictly stronger than the statement (2) Every real-indexed family of nonempty sets of reals has a choice ...
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Every Partially ordered set has a maximal independent subset

I am working on this problem:- Every Partially ordered set has a maximal independent subset. Definition:Let $\langle E,\prec\rangle$ be a partially ordered set. A subset $A\subset E$ is called ...
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Why in Teichmüller-Tukey lemma finitness is essential?

First we will state a Teichmüller-Tukey Lemma: Let $A$ be a set and $\phi$ be a property defined on all finite subset of $A$. Assume that $B$ is a subset of $A$ such that each finite subset of $B$ ...
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Finitely-additive measure over $\Bbb{N}$

On the set of natural number, we can consider the finitely-additive measure defined as: $$\mu(A) = \lim_{n\to\infty}\frac{\#(A\cap [1,n])}{n}.$$ However, there is a definable (by PA, or some ...
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Is the axiom of choice used here?

I was reading the following How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles? The answer by "user4594" basically considers a k-cell with rational ...
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One more AC equivalence question

Is "Every vector space admits a norm" weaker than AC? I know that the statement follows from "Every vector space has a basis", which is equivalent to AC.
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Example of a maximal idea

Let $A$ be the set of bounded continuous functions from the set of real numbers to itself. Then $A$ is a ring under pointwise addition and multiplication. The set $I$ of all functions $f \in A$ ...
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Question related to ordinal number without using Axiom of Choice.

Can we proof this result without using Axiom of Choice :- $$A\cap \alpha=\emptyset \,\,\,\, \mbox{and}\, \, \, A\times \alpha \sim A\cup \alpha$$ then there is an $A^{'} \subset A$ such that $\alpha ...
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Is this an accurate way to write the axiom of choice using transversals?

Is this an accurate way to write the axiom of choice using transversals? $\forall \mathcal{C} \exists T\, (T\subseteq \cup\,\mathcal{C} \land \forall X \exists s \forall t\, ((X\in\mathcal{C} \land ...
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Outer automorphisms of the infinite symmetric group

Denote by S$_\infty$ the group of permutations of $\mathbb N$. Question: Does there exist an outer automorphism of S$_\infty$, and if so, can one be exhibited? Does this depend on the continuum ...
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$Gal(\bar{\mathbb Q}/\mathbb Q)$ without choice, and constructive Galois theory

By this question: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice? We have that over ZF, algebraic closures of $\mathbb Q$ aren't unique. Are their Galois groups as extensions over ...
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Recursive use of the Axiom of Choice

In a standard proof that any sequence-compact metric space $(X,d)$ has a (finite) $\varepsilon$-net, the approach is the following: Make a sequence $(x_n)$ such that $$ x_{n+1}\notin\bigcup_{i=1}^n ...
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What is the name of proofs with (without) Axiom of Choice

In many contexts we distinguish between proofs using AC and proofs which do not use AC. (To phrase this somewhat differently: If there is a proof without AC, this proof is usually preferred.) I would ...
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Are countably infinite, compact, Hausdorff spaces necessarily second countable?

Let $X$ be a compact Hausdorff space. If $X$ is finite, then there are at most finitely many open sets and so it is second-countable. On the flipside if $X$ is uncountably infinite, it is possible ...
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The topological product of path-connected spaces is path-connected $\Rightarrow\sf AC$?

There is a very natural proof that the product of path-connected spaces is path-connected: Let $X=\prod_{i\in I}X_i$ be a product of path-connected spaces $X_i$. Given $(x_i)_{i\in I},(y_i)_{i\in ...
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Proving that axiom of choice is equivalent to this proposition

I'm trying to do the problem 8.2 of the book Notes on Set Theory of Moschovakis. I want to prove that Axiom of Choice is equivalent to this statement: $$\forall A\neq \emptyset,\, \forall ...
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A descending chain and choice axiom.

Let $R$ be a relation on a set $A$ such that for every $x\in A$ there exist a element $y\in A$ such that $x\mathrel{R}y$ the there exist a function $f\colon\omega\rightarrow A$ such that ...
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Is this proof using the axiom of choice correct?

In Munkres' Topology, the axiom of choice is defined as follows: Given a collection $A$ of disjoint nonempty sets, there exists a set $C$ containing exactly element from each element of $A$; that ...
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Choice-less Set Theory for Dummies

In almost every graduate set theory text there are some parts about equivalences of $AC$, its consequences, some axioms like $AD$ which imply $\neg AC$, some well-known axiomatic systems which $AC$ ...
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Construction of injective hulls without axiom of choice

Motivation: It is known that without the axiom of choice (AC), it is not provable that all categories of modules have enough injectives, let alone injective hulls. Still, there are examples of rings ...
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Do I need axiom of choice in this question?

If $X$ is a finite set and $Y$ an infinite one, I'm trying to prove there is a injective function $f:X\to Y$. My solution: Let $X=\{x_1,\ldots,x_n\}$, for each $x_i$ choose $y_1,\ldots,y_n$ such that ...
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Why is Zorn's Lemma called a lemma?

By convention a Lemma is a technical intermediate step which has no standing as an independent result. Lemmas are only used to chop big proofs into handy pieces. (Quoted from here) I am wondering why ...
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if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$ [duplicate]

I'm trying to prove if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$. I know this intuitively, since $f$ is surjective, for every $y$ in $Y$ we need just to choice an ...
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Proving statements , that require Zorn's lemma , for countable case directly by well-ordering principle of natural numbers

We know that for countable sets , the existence of a choice function is a consequence of the well-ordering principle ; and it is also known that the results like "every vector space has a maximal ...
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How do I prove that $S_A\cong S_B\implies |A|=|B|$?

Let $A,B$ be infinite sets such that $S_A\cong S_B$. (Symmetric groups are group isomorphic) How do I prove that $|A|=|B|$? The only proof I know uses Axiom of choice. (That is, using AC to give ...
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Additive function and continuity at a point

Does continuity at a point and Additive function imply continuity at all other points in a normed linear space. Is there some result like there exist a in field such that f(x) = ax for all x in normed ...
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Basic definition of continuity [duplicate]

Ltf(c+h) = f(c)(h goes to 0) if and only if Ltf(x) = f(c)(x goes to c). I am able to prove this fact using sequential criterion of continuity. But sequential criterion is dependent on Axiom of ...
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Open set in $\mathbb{R}$ as countable union of open intervals and which version of Choice [duplicate]

In proving, every non-empty open set in $\mathbb{R}$ is union of a countable collection of disjoint open intervals in $\mathbb{R}$. It seems to me this result is using some version of Choice(probably ...
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$\omega_2$ is not a countable union of countable sets [duplicate]

Without using axiom of choice, can we show that $\omega_2$ is not a countable union of countable sets? I know this cannot be done for $\omega_1$.
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Is $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q) = 0$ possible without choice?

That divisible abelian groups are precisely the injective groups is equivalent to choice; indeed, there are some models of ZF with no injective groups at all. Now, given that $\Bbb Q$ is injective, ...
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Proof that the finite product of nonempty sets is nonempty without axiom of choice from ZF

How do you prove that for $X_{i} \neq \emptyset$, $i \in \{1,...,n\}$ that $\prod_{i=1}^{n} X_{i} \neq \emptyset$ only using the ZF axioms but not the Axiom of Choice? I would like to see a rigorous ...
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Injections from all ordinals into a set $X$

We are working in $\mathsf{ZF}$. Let $X$ be a set. Let $A$ be the class of all injections $f: \alpha \to X$ for arbitrary ordinals $\alpha$. I am quite sure that, in fact, $A$ is a set, since if ...
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Does $A\times A\cong B\times B$ imply $A\cong B$?

This is similar to What does it take to divide by $2$? about $(A\sqcup A\cong B\sqcup B)\Rightarrow A\cong B$ which is valid in $\textsf{ZFC}$ by using cardinalities and also in $\textsf{ZF}$ by some ...
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Counterexample to the Hausdorff Maximal Principle

The Hausdorff Maximal Principle states: Every partially ordered set $\left(X,\leqslant\right)$ has a linearly ordered subset $\left(E,\leqslant\right)$ such that no subset of $X$ that properly ...
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Tychonoff principle iff Wellordering principle

Tychonoff Principle: Let $X_i$ for $i\in I$ be any sequence of non-empty sets indexed by the set $I$. Then the direct product $\prod_{i\in I}X_i$ is not empty, where $\prod_{i\in I}X_i$ is defined to ...
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Do an infinite set and its double have the same cardinality? [duplicate]

My question was inspired by this answer. Suppose $A$ is an infinite set. Does its double, $A\times\{0,1\}$, always have the same cardinality? In my head I quickly spotted a simple proof that the ...
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84 views

General Distributive Law and Axiom of Choice

Where can I find the proof of the fact that general distributive law of union over intersection and intersection over union is equivalent to Axiom of Choice? The mathematical formulation of the ...
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Does $\aleph_0\cdot\kappa=\kappa$ for every $\kappa\ge\aleph_0$ hold in ZF?

It is easy to show that for any (Dedekind) infinite cardinal $\kappa$ we have $\aleph_0+\kappa=\kappa$. Definition of an infinite cardinal is a cardinal such that $\aleph_0\le\kappa$. (I believe ...
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Prove that Open Sets in $\mathbb{R}$ are The Disjoint Union of Open Intervals Without the Axioms of Choice

There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?
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Prove that $\forall\alpha\geq\omega$, $|L_\alpha|=|\alpha|$ without AC

Without using Axiom of Choice, prove that $$\forall\alpha\geq\omega,~~|L_\alpha|=|\alpha|,$$ in which $\alpha$ is an ordinals, $\omega$ is the set of natural numbers, $L_\alpha$ is the $\alpha$-th ...
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Are there logical arguments against modern $\sf ZFC$ set theory?

As of asking this question, my knowledge of set theory is quite pedestrian. I've read about it in numerous nontechnical papers and even worked through three chapters of Jech - Set Theory, but in terms ...
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Choice-like axioms close to AC & rank-into-rank hypothesizes

Consider the following folklore theorems within ZF, Theorem 1: $V=L$ implies "$0^{\sharp}$ doesn't exist". Theorem 2: $V=L[U]$ implies "$0^{\dagger}$ doesn't exist". Theorem 3: $AC$ ...
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Axiom of Choice: What exactly is a choice, and when and why is it needed?

I'm having trouble understanding the necessity of the Axiom of Choice. Given a set of non-empty subsets, what is the necessity of a function that picks out one element from each of those subsets? For ...
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Statements comparable with Axiom of Choice in ZF

Let $AC$ denote any fixed statement of the Axiom of Choice in $ZF$. Consider the set of statements $\phi$ in the language of $ZF$ such that either $ZF+\phi$ proves $AC$ or $ZF+AC$ proves $\phi$. The ...
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Adjunctions via Reflections and the Axiom of Choice

I have met two ways of defining adjunctions: via the triangle identities, and via reflections. Proposition 3.1.2 Let $F:\mathsf A \rightarrow \mathsf B$ be a functor and $B$ an object of $\mathsf ...
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Existence of finite sets of infinite set without using AC

Is it possible to prove that every infinite set $B$ has a subset of cardinality $n$, for every natural $n$, without using AC? I know how to prove this claim by induction. In the induction step I chose ...