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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reverse Banach-Tarski paradox

Is it possible to reverse the operation of the Banach-Tarski paradox? That is, I have two three-dimensional balls, and is it possible to combine them into one ball that is identical to one of the ...
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3answers
634 views

Banach-Tarski theorem without axiom of choice

Is it possible to prove the infamous Banach-Tarski theorem without using the Axiom of Choice? I have never seen a proof which refutes this claim.
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Doesn't axiom of choice only apply to countable many groups?

The process of selecting an object from each group seems to be similar to counting. It doesn't make sense to apply it to uncountable many groups.
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3answers
545 views

Advantage of accepting non-measurable sets

What would be the advantage of accepting non-measurable sets? I personally feel that non-measurable sets only exist because of infamous Banach-Tarski paradox...
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2answers
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(Revisited$_2$) Injectivity Relies on The Existence of an Onto Function Mapping Back to Its Preimage

QUEST: For any sets $X$ and $Y$, there exists an injective function $f:X\rightarrow Y$ if and only if there exists a surjective function $g:Y\rightarrow X$. QUESTION$_1$: How do you people ...
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1answer
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Zorn's lemma converse? (Context: Maximal proper subgroups)

So, in my qual prep class a pretty simple question popped up: "Prove that for any nontrivial finite group there exists a maximal proper subgroup." So of course, my natural inclination was to ...
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1answer
45 views

$|\alpha\times\alpha|=|\alpha|$ for infinite ordinal $\alpha$, definably

I am trying to prove Proposition 1.7 of http://math.bu.edu/people/aki/7.pdf: Proposition 1.7 (Halbeisen–Shelah). If $\aleph_0\le|X|$, then $|{\cal P}(X)|\not\le|{\rm Seq}(X)|$. In words, ...
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3answers
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What's wrong with this proof of the inconsistency of the axiom of choice?

Let $\mathscr{T}$ be the (countable) collection of all theorems provable in ZFC. Define an equivalence relation on $\mathscr{T}$ by $\phi\sim\psi$ iff $(\phi \iff \psi)$. In other words, two theorems ...
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1answer
25 views

Choice function for collection of arbitrary finite sets. AC required?

I understand how we can show the existence of a choice function for any (finite or infinite) collection of (finite or infinite) subsets of, say, $\mathbb{N}$ or $\mathbb{Z}$ without using the axiom of ...
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1answer
28 views

Looking for explanation of Banach-Tarski Proof, preferably by visual methods “Video, Pictures, Diagrams…”

could someone please explain the four steps of Banach-Tarski? 1- Find a paradoxical decomposition of the free group in two generators. 2- Find a group of rotations in 3-d space isomorphic to the free ...
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0answers
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Strength of “Every finite dimensional subspace of a vector space has a complement”

Does the following choice principle have a name? Every finite dimensional subspace of a vector space has a complement. Equivalently, every line inside a vector space has a complementary ...
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1answer
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Is AC necessary to show that in metric spaces $x\in\operatorname{closure}(A)$ implies $\exists\{a_n\}_{n=1}^\infty\subseteq A$ s.t. $\lim a_n=x$?

Let $(X,d)$ be a metric space. Let $x\in\operatorname{closure}(A)$, where $A\subseteq X$. Then for each $n\in\mathbb{N},\exists x_n\in B_{\frac{1}{n}}(x)\cap A$, where $B_\varepsilon(x)$ is the open ...
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1answer
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If $|A|<|B|$ does $B$ surject onto $\aleph(A)$?

After reading Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC I became curious if there is a generalization to arbitrary cardinals. That is, if $\frak m<n$, does it follow ...
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1answer
42 views

How to show countability of $\omega^\omega$ or $\epsilon_0$ in ZF?

I know that with choice, the countable union of countable sets is countable, making $\omega^\omega$ and $\epsilon_0$ both countable. Can we show this without choice? E.g. in the case that $\omega_1$ ...
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1answer
73 views

How to prove that $\sf CH$ implies $2^{\aleph_0}=\aleph_1$

Of course, most of you will, upon reading the title, exclaim "But isn't that the definition of the continuum hypothesis?" So I need to be a little more careful about the exact definitions. Let ...
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1answer
66 views

Can we prove, without axiom of choice, that the set of all zero divisors (including $0$) of a commutative ring with unity contains a prime ideal?

Let $R$ be a commutative ring with unity , I know that assuming axiom of choice , if $A$ is the set of all zero divisors (including $0$ ) then it is a union of prime ideals so it contains a prime ...
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0answers
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Does “$(\exists f:A\twoheadrightarrow B)\implies(\exists f:B\hookrightarrow A)$” implies the axiom of choice? [duplicate]

Let $P$ denotes the property that if there exists a surjection from set $A$ to set $B$, then there exists an injection from $B$ to $A$. It's apparent that $P$ can be proved in ZFC. My question is that ...
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1answer
18 views

The existence of the sequence corresponding to some asymptotic sequence

The following proof of the axiom of choice by induction is obviously false: Let $(\Lambda)_{i=1, 2, \ldots}$ be an infinite sequence of nonempty sets. When $i=1$, self-evident. We will assume this ...
2
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2answers
33 views

Does a direct sum decomposition of an infinite-dimensional vector space require Zorn's lemma?

Let $V$ be an infinite-dimensional vector space and $V'\subset V$ a subspace. Does it require Zorn's lemma to write $V=V'\oplus V''$ for some other subspace $V''\subset V$?
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2answers
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Axiom of choice : continuous function and uniformly continuous

How I proof that every continuous function f in [0,1] is uniformly continuous, without axiom o choice? I took this from the book Axiom of Choice from Horst Herrilich He had a observation that ...
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2answers
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Does the proof of Bolzano-Weierstrass theorem require axiom of choice?

When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of ...
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1answer
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Can we prove AC from the statement “There is no $\aleph$ cardinal strictly between $\operatorname{CARD}(X)$ and $\operatorname{CARD}(2^X)$”?

If $X$ is a set, let $\operatorname{CARD}(X)$ denote the Cardinal number of $X$. Let GCH(1) be the statement "If $K$ is an infinite initial ordinal number, then there exists no initial ordinal number ...
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1answer
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Is it consistent without the axiom of choice that every permutation of some infinite set have fixed points?

A "permutation" of a non-empty set means an injective mapping of the set onto itself. Let $S(1)$ be the statement "There exists a permutation of every set containing at least two elements, which has ...
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2answers
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Choosing a Cauchy sequence for a real

It is easy to form in ZF, for each real $a$, a "canonical" Cauchy sequence that converges to $a$. For example, one can take the sequence of finite initial segments of the decimal expansion of $a$, ...
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1answer
44 views

A question about infinite sets and Cantor's Power Set theorem

Let $\operatorname{Card}(X)$ denote the cardinal number of the set $X$. The standard proof of Cantor's Power Set theorem stating that "$\operatorname{Card}(X) < \operatorname{Card}(2^X)$" is ...
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1answer
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Axiom of choice for singletons

Let $\mathscr C \subseteq \mathscr P (\mathbb R )$ be a family of singletons, i.e.: each element of $\mathscr C$ contains exactly one real number. Let $f: \mathscr C \to \mathbb R $ be the function ...
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1answer
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Zorn's Lemma's chain condition

Zorn's Lemma requires that every chain in a partially ordered set $X$ has an upper bound. In this article Gowers uses Zorn's Lemma to find a maximal linearly independent (over $\mathbb{Q}$) subset of ...
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1answer
30 views

Regularity of $\omega_1$ and axiom of choice

Why is the regularity of the ordinal $\omega_1$ a consequence of the axiom of choice?
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50 views

Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

Does the ring of formal power series $\mathbb{F}[[x]]$ as a vector space over $\mathbb{F}$ admit a basis without assuming the Axiom of choice, at least in some special cases of $\mathbb{F}$? I'm ...
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1answer
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Breaking the AC barrier using Kelley-Morse set theory

In her blog post Variants of Kelley-Morse set theory, Prof. Gitman proves that every model of the the common version of Kelley-Morse set theory ($\mathsf{KM}$) is a model of the Wikipedia version of ...
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1answer
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In P. Cohen's models (or others) may we have $\neg\mathsf{AC}+\mathsf{CH}$? May we have $\neg \mathsf{AC} + \neg \mathsf{CH}$?

I know we have the consistency of $\mathsf{ZF} + \mathsf{AC} + \mathsf{GCH}$, $\mathsf{ZF} + \neg \mathsf{AC}$, and $\mathsf{ZF} + \mathsf{AC} + \neg \mathsf{GCH}$. What about $\mathsf{ZF} + \neg ...
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1answer
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Difference between Zorn's Lemma and the ascending chain condition

Let $S$ be a non-empty partially ordered set with respect to a relation $\leq$. Then: Zorn's Lemma: If $S$ has the property that any totally ordered subset $U\subset S$ has an upper bound, then ...
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Choice of a skeleton

Suppose we are in presence of a strong enough axiom of choice (e.g., choice for conglomerates). I know that any category has a skeleton, but I would like to know if I can choose a skeleton which ...
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2answers
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Is the axiom of choice really all that important?

According to this book: The Axiom of Choice is the most controversial axiom in the entire history of mathematics. Yet it remains a crucial assumption not only in set theory but equally in modern ...
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7answers
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Can you explain the “Axiom of choice” in simple terms?

As I'm sure many of you do, I read the XKCD webcomic regularly. The most recent one involves a joke about the Axiom of Choice, which I didn't get. I went to Wikipedia to see what the Axiom of ...
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3answers
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Equivalence of countable choice for subsets of the reals and “second countable $\implies$ Lindelöf”

Looking for a proof that in a second countable space every open cover has a countable subcover -- i.e. every s.c. space is a Lindelöf space -- I bumped into this question. That answered my question. I ...
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2answers
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Prove there exists a strictly increasing function from the natural numbers to a partially strictly ordered set

Let $P$ be a non-empty partially strictly ordered set and assume no element of $P$ is maximal (i.e. for every $x \in P$ there exists $y\in P$ with $x < y$). Show there exists a function ...
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0answers
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Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
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Every II-finite set is III-finite

I need some help proving that if a set $X$ is II-finite then it is III-finite, i.e. if every non-empty family of subsets of $X$ which is linearly ordered by inclusion has a maximal element under ...
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1answer
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Lebesgue measure without choice

From this question and this question (and their answers) I gather that it is consistent with ZF without The Axiom of Choice to assume that there exist countable sets $A_n$, $n\in \mathbb N$, such that ...
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1answer
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Accumulation point is a limit of some sequences in first-countable space with Axiom of Choice.

I have a statement which says that Let $X$ be a first-countable space and $E$ be a subset of $X$. If $p$ is an accumulation point of $E$, then there exists a sequence in $E$ that converges to ...
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1answer
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Finite complement topology over $\Bbb{R}$ is not second-countable under ZF?

Under $\mathsf{ZF+AC_\omega}$, the space $\Bbb{R}$ with finite complement topology is not second-countable, since a countable union of countable subsets of $\Bbb{R}$. However, such argument doesn't ...
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1answer
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Without AC, it is consistent that there is a function with domain $\mathbb{R}$ whose range has cardinality strictly larger than that of $\mathbb{R}$?

I stumbled across this question earlier, and the top comment on the bottom answer asserts two claims: Without the Axiom of Choice, It is consistent that there exists a function with domain ...
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1answer
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Every set with more than point admits a permutation with no fixed point and the Axiom of Choice [duplicate]

Assuming axiom of choice , for any set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ . Is the converse true , i.e. Does the statement " ...
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2answers
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Isomorphic Free Groups and the Axiom of Choice

When I read about free group, the proof which concerns about two free groups $F(X)$ and $F(Y)$ are isomorphic only if $\operatorname{card}(X) = \operatorname{card}(Y)$ has a sentence going as follows: ...
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1answer
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Consequences of the negation of the Axiom of Dependent Choice

It seems to me that a proper reason to include The Axiom of Choice as a foundational axiom of set theory should be based on the observation that the negation of The Axiom of Choice has absurd ...
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1answer
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In any commutative ring with unity, every prime ideal is finitely generated implies every ideal is finitely generated; can it be prove without A.C.?

Assuming Zorn's lemma, "In any commutative ring with unity, if every prime ideal is finitely generated, then every ideal is finitely generated". Is the converse true, i.e. if in any commutative ring ...
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1answer
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In every ring with unity satisfying ACC, every ideal is finitely generated; can we prove it without assuming Axiom of choice?

Assuming Zorn's lemma, we can prove that in every ring with unity satisfying Ascending-Chain-Condition, every ideal is finitely generated. Is this statement equivalent to Zorn's lemma? Can we prove it ...
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2answers
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$\forall \alpha \exists \beta: \beta > \alpha$ where $\alpha$ and $\beta$ cardinals

I have to prove ZF $\vdash$ $\forall \alpha \exists \beta:\beta > \alpha$, where $\alpha, \beta-$ cardinal numbers. I can prove it only in ZFC. Let's fix some cardinal number $\alpha$. By ...
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1answer
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Assuming the axiom of choice ,how to prove that every uncountable abelian group must have an uncountable proper subgroup?

Assuming the axiom of choice , how to prove that every uncountable abelian group must have an uncountable proper subgroup ? Related to Does there exist any uncountable group , every proper subgroup ...