Questions tagged [axiom-of-choice]

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. Use this tag in tandem with (set-theory).

The axiom of choice is an axiom usually added to the Zermelo–Fraenkel set theory (ZF) stating that given a set $I$ and $a_i$, for $i\in I$, non-empty sets there exists a function $f\colon I\to \bigcup a_i$ such that $f(i)\in a_i$ such a function is called a choice function since it chooses one element $f(i)$ from each of the sets $a_i$.

It is equivalent to the statement that every set can be well-ordered, as well to Zorn's lemma which asserts that if a partial order has the property that every chain is bounded from above, then there is a maximal element.

The axiom of choice is generally independent of ZF, and if ZF is consistent then it is consistent with both the axiom of choice as well its negation.

Some theorems which follow from the axiom of choice:

  1. The product of compact spaces is compact (equivalent)
  2. Every surjective map has a right inverse (equivalent)
  3. Every vector space has a basis (equivalent)
  4. Countable union of countable sets is countable
  5. Every infinite set has a countable subset
  6. Every field has an algebraic closure
  7. There are sets of real numbers which are not Lebesgue measurable

Most of the mathematicians nowadays assume the axiom of choice when they deal with their mathematics. Mostly because without it infinite processes in mathematics become much harder to handle, and one can construct models without the axiom of choice in which continuity of sequences is not equivalent to $\varepsilon$-$\delta$ continuity, or some fields have no algebraic closure.

See also:

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Why is the Axiom of Choice not needed when the collection of sets is finite?

According to Wikipedia: Informally put, the axiom of choice says that given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin. In many cases such a selection…
Juan
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Does choosing a counting function require the Axiom of Choice?

Say one is asked to prove something about set $X$, and part of the given conditions is that $X$ is countable. Your proof starts with "Let $f: X \mapsto \Bbb{N}$ be a counting function for $X$ and ..." and the proof proceed to use $f$ to…
Mark Fischler
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Does 'let x be a member of S...' require axiom of choice?

A lot of proofs start like this 'Let x be some member of the set S, then...' (If S are the natural numbers, this is especially common) Do all of those proofs essentially require axiom of choice? Or is it possible to talk about an element of a set…
Thomas Ahle
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Does a set become a proper class with $AC$ (or vice versa)?

I hope the following questions are not fundamentally stupid to raise. Let A be a class term in the form of $ A = \{ x | \phi \}$. Claim $1$: $A$ is a proper class in ZFC if and only if it is also a proper class in ZF. As I try to prove this, this…
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Countable union of countable sets is countable and AC (or axiom of countable choice)

It is written in tag page of Axiom of choice in MSE that Countable union of countable sets is countable is a theorem which follows from AC. Do we really need AC to prove this? Please see following proof from Real Analysis by Bartle and Sherbert. I…
Sushil
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Zorn's lemma on proper class with chains bounded in cardinality

Basically, I'm trying to understand a note in Jacob Lurie's paper On a conjecture of Conway (Illinois Journal of Mathematics 46.2, 2002). Let $X\subseteq Y$ be sets, $\mathbf{U}$ a proper class, and $\varphi\colon X \to \mathbf{U}$ an injective…
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Do I need axiom of choice to construct $\sqrt x$?

I had a lecture concerning axiom of choice, and the teacher says that an equivalent property is that surjective maps are invertible at right. After, as an example, we had the function $\sqrt x$. The example goes as follow : $x\mapsto x^2$ is…
user659895
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When and how to use Zorn's lemma?

I was reading the proof of the fact that every nonzero proper ideal in a dedekink domain factors uniquely into a product of prime ideals. I was stumped by the beautiful application of Zorns Lemma to this prove theorem. It didn't even occur to me…
Mohan
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Without axiom of choice

We denote by $A$, $A′$, $B$, $B′$ the four sets. We are given mapping $u:A′→A$, $v:B→B′$. We denote by $F(A, B)$ the set maps $A→B$ ,$F(A′, B′)$ the set maps $A′→B′$. Let $Φ : F(A, B) → F(A′, B′)$ be a mappig which makes $v◦f ◦ u ∈ F(A′, B′)$…
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Existence of a sequence of elements picked arbitrarly

How to prove with the axiom of choice that : Given a family of non empty sets $(A_n)_{n\in\mathbb N}$, there exists a sequence $(x_n)_{n\in\mathbb N}$ such that for any $n\in\mathbb N$, $x_n \in A_n$. I don't know how to proceed... fixing $n$ then…
Célestin
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Axiom of Choice : Is it Axiom? or Trivial Fact?

This question might be rude or inappropriate, but I am very beginner of studying mathematics. Today I had learned the concept of Axiom of Choice, which states that $ \forall$ nonempty indexed set $S_i$, there $\exists$ indexed family of element…
Daschin
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Can we do this kind of choice in ZF?

I'm trying to see if the following proof use any kind of choice beyond the choices ZF allows: Assume nested intervals theorem on real numbers. So shortly, if $I_n$ is a sequence of nested intervals whose length converges to $0$ (with closed…
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Is there some weaker form of choice involved here?

Consider the following partition of $\mathbb{Q}^2$ into disjoint classes : $(p_1, p_2) \sim (q_1,q_2)$ if and only if both $p_1 - q_1$ and $p_2 - q_2$ have odds denominators when written in their lowest terms. Suppose I endow the equivalence class…
M.G
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Open set in $\mathbb{R}$ as countable union of open intervals and which version of Choice

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 AC). Am I correct is it using some version of…
Sushil
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How does $\neg$AC lead to a partition of the reals with more parts than reals?

I heard a while ago that the negation of the axiom of choice leads to the existence of a partition of the real numbers, into more partitions then there are real numbers. I found this post which says You don't want non-measurable sets? This means…
Numeral
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