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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Axiom of Choice: Where is falsity in argument and understanding to put is as Axiom?

In terms of purely set theory, the axiom of choice says that for any set $A$, its power set (with empty set removed) has a choice function, i.e. there exists a function $f\colon \mathcal{P}^*(A)\...
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2answers
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Polynomial ring with arbitrarily many variables in ZF

For a given field $k$ and a set $X$ we want to define the ring $k[X]$ of polynomials with $X$ as the set of variables. We do not assume $X$ to be finite. And we want to do this without employing axiom ...
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1answer
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Why is the axiom of choice controversial? [duplicate]

In other words, what are the arguments for ZF over ZFC, and what philosophical issues have people raised against including it as a standard axiom of set theory?
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1answer
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The cardinality of a countable union of countable sets, without the axiom of choice

One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have ...
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2answers
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Existence of model of ZF in which every uncountable cardinal is the cardinality of some power set?

Does there exist a model of ZF in which any uncountable cardinal number is equal to the cardinality of the power set of some set ? ( In ZFC it is not possible as is shown by the answers to this ...
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1answer
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When do minimal subcovers always exist - without choice?

In my answer to Doubt in the definition of a compact set, I sketched a proof of the following fact: Suppose $X$ is a topological space such that every open cover of $X$ has a minimal subcover. ...
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1answer
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An explicit construction for an everywhere discontinuous real function with $F((a+b)/2)\leq(F(a)+F(b))/2$?

I would like to know an explicit method on constructing an everywhere discontinuous real function $F$ with the property: $$F((a+b)/2)\leq(F(a)+F(b))/2.$$ There is a non-constructive example (with the ...
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1answer
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Tychonoff's theorem for completely regular spaces and the axiom of choice

It is well-known that Tychonoff's theorem, i.e., that the product of any set-indexed family of compact spaces is compact, is equivalent to the axiom of choice. It is also the case that if the spaces ...
2
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2answers
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Sum of measurable functions is measurable: countable choice required?

The standard proof that the sum of measurable functions is measurable uses countable choice, via the countable subadditivity of outer measure ($\implies$ measurable sets are closed under countable ...
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2answers
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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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4answers
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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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Existence in variants of the Axiom of Choice

Let $\{A_i: i \in I\}$ be a nonempty family of nonempty sets. Why is it allowed to prove the Axiom of Choice using the Well Ordering Principle as follows: There is a well-ordering of $\cup_{i \in ...
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Stochastic differential equation with weak solution for which pathwise uniqueness hold: is the axiom of choice necessary?

A stochastic differential equation with a weak solution for which pathwise uniqueness hold has a strong solution (results by Yamada and Watanabe (1971) on the weak and strong solutions). Does the ...
2
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1answer
31 views

If $a\sqcup b$ and $a\times b$ biject, then $b$ either injects or surjects in-/onto $a$

Let $a$ and $b$ be sets such that there is a bijection $a\sqcup b\to a\times b$. Show, without assuming the Axiom of Choice, that there is either a surjection $b\to a$ or an injection $b\to a$. ...
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1answer
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A $\sigma$-algebra equivalent to the Lebesgue algebra under CC

Working in $\sf ZF$, is there a natural definition of an algebra $\Sigma$ with the following properties: $\Sigma$ is a $\sigma$-algebra on $\Bbb R$, i.e. it is closed under complement and countable ...
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2answers
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Axiom of Choice and LEFT inverse [duplicate]

I am aware of why the Axiom of Choice is equivalent to the the statement that every surjection splits. However, I don't see why we don't also need AC to show that every injection splits. In ...
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0answers
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A classification of the types of poset extensions that raise local-global principle is equivelent to the axiom of choice?

EDIT: Completely overhauled the question, remove category theory from it (stated in plain set theory): The original question Adding relations to a partial order Let $X$ be a set and let $P: Pow(X) \...
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1answer
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Proving cardinality of coproduct presentation is unique without choice?

The definition of an extensive category immediately implies that given two coproduct decompositions indexed by sets of equal cardinality, if the coproduct objects are isomorphic compatibly with their ...
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4answers
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Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?

I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of ...
10
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4answers
356 views

Countable choice and term extraction

The constructive Axiom of Countable Choice (ACC) is widely accepted due to its computational content. It states that: $$ \forall n\in \mathbb{N} . \exists x \in X . \varphi [n, x] \implies \exists f: ...
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1answer
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Can the unit interval be the disjoint union of countably many “super-dense” parts?

I'm curious about this question in the case where $f$ is not necessarily measurable. I think what it comes down to is this: Is there an $\varepsilon<1$ and a partition of $[0,1]$ in countably ...
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3answers
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What is an example of two sets which cannot be compared?

In set theory, if we do not assume the Axiom of Choice, we cannot prove the Trichotomy Law between cardinals. That is, we cannot prove that for any two sets, there exists an injection from one to the ...
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2answers
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No uncountable ordinals without the axiom of choice?

In Uncountable ordinals without power set axiom Francois Dorais explains that without the Power-set Axiom we cannot prove the existence of uncountable ordinals. I am guess that the power set of an ...
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Original proof of Zorn's Lemma

On the wiki page for Zorn's Lemma it says that this lemma was Proved by Kuratowski in 1922 and independently by Zorn in 1935 but then it says: Zorn's lemma is equivalent to the well-ordering ...
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1answer
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Need this transfinite construction reach a closed set in the absence of the Axiom of Choice? Or Hartogs' theorem?

In this question I'm primarily concerned with the workings of a set theory that lacks both Foundation and Choice. Given a set $X$ and a function $\sigma:\mathcal{P}(X)\to\mathcal{P}(X)$, we have a ...
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2answers
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Why do people accept the axiom of choice given the well ordering principle?

We know without any doubt that the axiom of choice implies (in fact is equivalent to) the well ordering principle. The well ordering principle can't be true! If we take the open interval $(0,1)$ for ...
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2answers
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Is there a known well ordering of the reals?

So, from what I understand, the axiom of choice is equivalent to the claim that every set can be well ordered. A set is well ordered by a relation, $R$ , if every subset has a least element. My ...
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1answer
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Soft Question: Why does the Axiom of Choice lead to the weirdest constructions?

I hope this is not too off-topic / soft for math.stackexchange. My basic question is: why does the Axiom of Choice allow for some of the weirdest constructions in math? I'll make a list of the weird ...
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1answer
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Are there any large cardinals that are not ordinals? [closed]

In ZF, are there any useful large cardinal that cannot be well-ordered? I think that some of the partition cardinals are that way, since with AC, we cannot have $\kappa \to (\omega)^{\omega}$. Are ...
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1answer
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Non-principal ultrafilters on a set [duplicate]

Let $X$ be a set. If $X$ is finite then all ultrafilters on $X$ are principal, i.e. have the form $\{A \subseteq X : x \in A\}$ for some $x\in X$. But now suppose $X$ is infinite, say $X=\mathbb N$. ...
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1answer
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Axiom of Choice implies the Well-Ordering Principle

I am trying to understand the proof of this implication we were taught in my set theory module. I cannot seem to tie it together with the final line of the argument... We used this lemma: Given $F\...
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2answers
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Can every poset be toplogically sorted?

A partial ordering $\preceq'$ on $S$ is said to be compatible with another partial ordering $\preceq$ if for all $a,b,\in S$, $$ a \preceq b \Rightarrow a \preceq' b$$ Given a poset $(S, \preceq)$, ...
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1answer
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The well-ordering principle implies Zorn's Lemma

I have read and understood proofs for each implication between $AC$, $ZL$, $WO$ except this one. These proofs need about 10 lines each. Can someone share a neat, hopefully short, proof for $WO\implies ...
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1answer
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A puzzle concerning the axiom of choice and the reals

Recently I was told the following riddle: Let $A=(a_1,...a_n,...a_{2n},a_{2n+1})$ a 2n+1-tuple of real numbers with the following property: Whatever number $a_i$ is removed from $A$ the remaining 2n ...
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1answer
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Proving the Axiom of Choice is equalivalent to the statement “If $A$ can be well-ordered, then so can $\mathcal{P}(A)$.”

I am still not completely confident in proving equivalence between the Axiom of Choice and statements such as the one posed in the title, so I want to make sure that I am on the right track. So ...
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Existence of how many sets is asserted by the axiom of choice in this case?

Applying the axiom of choice to $\{\{1,2\}, \{3,4\}, \{5,6\},\ldots\}$, does only one choice set necessarily exist, or all of the $2^{\aleph_0}$ I "could have" chosen? Or something in between? It ...
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1answer
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Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly

While reading David Williams's "Probability with Martingales", the following statement caught my fancy: Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel $\sigma$-...
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1answer
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Does the regularity of $\omega_{\alpha+1}$ need Axiom of Choice?

Many books indicate yes to this question. However, I found the only lemma they claim to use AC is the following statement: If $\{A_i\}_{i\in I}$ is a family of sets, then $|\bigcup_{i\in I}A_i|\leq|I|...
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1answer
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For every cardinal $\kappa$, $\kappa^+$ is regular

Again I'm struggling with a proof from this introduction to cardinals. Lemma 2.6. For every cardinal $\kappa$, $\kappa^+$ is regular. Proof. If not, then there would be a cofinal map $f:\...
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1answer
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Using the axiom of choice to choose bijections

I couldn't think of a better question title. I am trying to understand the proof of theorem 1.8 in this introduction to cardinals. Theorem 1.8. Let $\kappa\in CARD$. Let $X=\bigcup_{\alpha<\...
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1answer
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Regarding Choice in fields outside set theory.

When authors say stuff like The equivalence of continuity and sequential continuity in metric spaces uses(/requires) some version of the axiom of choice. Are they assuming that we are working ...
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Maximal ideals in $C(X)$ and Axiom of Choice

The following result are true if we assume full axiom of choice: A. If $X$ is a compact Hausdorff space, then every maximal ideal of the ring $C(X)$ has the form $A_p=\{f\in C(X); f(p)=0\}$. B. If $...
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1answer
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Without the Axiom of Choice, $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$?

Question: In ZF (so AC does not necessarily hold) does the following claim hold? $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$ This question arose to me when reading the top ...
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1answer
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Comparability of a set and a subset of power set.

It's well known that for any set $A$, $A < P(A)$. But now, I have some question that, WITHOUT AC, can we guarantee that $A \leq X$ or $X \leq A$ whenever $X \subseteq P(A)$? Thank you.
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1answer
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Does König's inequality require axiom of choice?

König's inequality: If $\kappa_\alpha<\lambda_\alpha$ for all $\alpha<\mu$, then $\sum\kappa_\alpha<\prod\lambda_\alpha$. In order to prove this inequality, we need to provide an injection ...
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Is the equivalence of the ascending chain condition and the maximum condition equivalent to the axiom of dependent choice?

Assuming the axiom of dependent choice, for a partially ordered set $(X,\le)$, the following statements are equivalent: $X$ fulfils the ascending chain condition, i. e. every chain $x_1\le x_2\le\...
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1answer
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Does the relationship between Jonsson Cardinals and Jonsson Algebras require the axiom of choice?

Using Skolem functions, one can see in ZFC that a cardinal $\kappa$ is Jonsson iff there are no Jonsson algebras on $\kappa$. (I.e. every algebra of size $\kappa$ has a proper subalgebra of size $\...
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Proving Hahn-Banach theorem directly

Is is possible to find an example of nonreflexive Banach space for which one can establish Hahn-Banach theorem directly (not referring to the axiom of choice)?
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Hyperplanes without Axiom of Choice

For any projective space that contains more than one point, is it possible to prove that it contains a hyperplane without using the Axiom of Choice? It's easy enough to prove that there exists a ...
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1answer
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Hausdorff Maximal Principle

"Hausdorff's Maximal Principle" says that any partial order P has a maximal chain (chain = linear suborder). It is equivalent to the axiom of choice. If we restrict Hausdorff's Maximal Principle to ...