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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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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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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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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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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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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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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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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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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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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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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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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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 ...
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 ...
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 $\sigma$-...