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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1answer
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Is a vector space isomorphic to the kernel $\oplus$ image of a map out of it?

Let $f:V\to W$ be a linear map of finite-dimensional vector spaces. By simply counting dimensions and using rank-nullity, it is clear that $V\cong \mathrm{im}\,f\oplus\mathrm{ker}\,f$. I want to know ...
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Advantage of accepting the axiom of choice

What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)? It seems that there is no clear reason to prefer over other axioms.. Thanks for help.
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2answers
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Simple question: Which is the Wikipedia definition of axiom of choice

I looked up Wikipedia, and on the top of the page it says: For every indexed family $(S_i)_{i \in I}$ of nonempty sets, there exists an indexed family $(x_i)_{i \in I}$ of elements such that $...
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1answer
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Without AC is there a relationship between $\beth$ and $\aleph$ numbers?

Assuming AC we know that all $\beth_\alpha$'s will be $\aleph_\beta$ for some $\beta$ since they can be well ordered. Can anything interesting be said about their relationship without AC? Is it ...
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1answer
60 views

How to prove axiom of limitation of size with axiom of replacement and global axiom of choice [closed]

In the book<>written by Charles C.Pinter it said that in order to prove "the axiom of replacement together with strengthened version of axiom of choise imply axiom of limitation of size"(X is a ...
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1answer
41 views

Why is axiom of dependent choice necesary here? (noetherian space implies quasicompact)

In the proof of noetherian space implies quasicompact I am reading, it goes as follows: Let $X$ be a noetherian space. Let $U$ be the collection of open subsets of $X$ that can be expressed as a ...
0
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1answer
39 views

Proof of Ultrafilter lemma with two propositions and Zorn lemma

I would like to prove the following: Let $X$ be any set, then every filter $\mathcal{F}$ on $X$ is contained in an ultrafilter $F$ Using two propositions and Zorn Lemma. I am required to come ...
2
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0answers
78 views

Unable to prove that a statement is an equivalent form of the Axiom of Choice [duplicate]

The exercise 22 page 158 of Elements of Set Theory by B. Enderton is the following: Show that the following statement is another equivalent version of the axiom of choice: For any set $A$ there ...
3
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1answer
210 views

Is it possible to prove $|V_\kappa|=\kappa$ for strongly inaccessible $\kappa$ without AC?

First, let me note that my definition of "strongly inaccessible" is that a nonzero ordinal $\kappa$ is strongly inaccessible if ${\rm cf}(\kappa)=\kappa$ and $\forall\alpha<\kappa, {\cal P}(\alpha)\...
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2answers
377 views

Relationships between AC, Ultrafilter Lemma/BPIT, Non-measurable sets

How is it possible to reconcile the following... In 1970, Solovay constructed Solovay's model, which shows that it is consistent with standard set theory, excluding uncountable choice, that all ...
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2answers
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Lebesgue measurable but not Borel measurable

I'm trying to find a set which is Lebesgue measurable but not Borel measurable. So I was thinking of taking a Lebesgue set of measure zero and intersecting it with something so that the result is not ...
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2answers
284 views

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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1answer
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ZF and the Existence of Finitely additive measure on $\mathcal{P}(\mathbb{R})$

My understanding is that Solovay (1970)'s relative consistency shows that if ZFC+I has a model then ZF+DC has a model in which every subset of the reals is Lebesgue measurable (and hence $\sigma$-...
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4answers
818 views

Most functions are measurable

My professor once said that if you did not use the axiom of choice to build a function $f : \mathbb{R}^n \to \mathbb{R}^m$, then it is Lebesgue measurable. To what extent this is true?
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0answers
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Naive understanding of choice axiom [duplicate]

It's written in many resources that if we consider only finite sets, that choice axiom can be skipped and be proven from other ZF axioms. I remember I've read the following explanation. If $A$ is a ...
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4answers
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Axiom of Choice: Where does my argument for proving the axiom of choice fail? Help me understand why this is an axiom, and not a theorem.

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 ...
4
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1answer
52 views

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 ...
2
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2answers
76 views

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 ...
3
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1answer
70 views

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. ...
4
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1answer
189 views

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 ...
2
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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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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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7answers
549 views

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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1answer
32 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
34 views

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
49 views

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
915 views

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 ...
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4answers
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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
55 views

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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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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2answers
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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
59 views

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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2answers
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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 $\sigma$-...