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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Construction of a choice function on the open subset of R²

I am trying to construct a choice function on the open subset of $\mathbb{R}^2$ (with the topology induced by the euclidiean distance) I cannot use maximum and minimum because if the set is open I might not have these. My first try is to find a…
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definition of Axiom of Choice

The Cartesian product of any nonempty collection of nonempty sets is nonempty. In other words, if $I$ is any nonempty (indexing) set and $A_i$ is a nonempty set for all $i\in I$, then there exists a choice function from $I$ to $\cup_{i\in…
jk001
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Examples of two definitions equivalent in ZFC, but inequivalent in ZF set theory.

What are two definitions of an object or class of objects, which are equivalent assuming the axiom of choice, but are inequivalent assuming only ZF set theory? I am not looking for theorems that are inequivalent, more like interesting examples of…
user107952
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Why isn't this an axiom of choice?

The standard axiom of choice can be expressed in the first order language of ZF by $$\forall X\big((X\neq\emptyset\wedge\emptyset\notin X)\implies\exists f:X\to\bigcup X\ \forall Y\in X(f(Y)\in Y)\big).$$ Naively, a simpler version would seem to be…
Alec Rhea
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Is Axiom of Choice related to Indeterminism

Is it fundamentally wrong to think if the Axiom of Choice (AC) as a logical codification of indeterminism? Could it not be related to matching distributions with a random number generator?
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Is AC required to demonstrate that every collection of disjoint open intervals in R is countable?

I understand that to prove this, one considers that each open interval contains a rational number and since the open intervals are disjoint, there exists an injection from the collection of intervals to the rational numbers, which are countable,…
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An application of the axiom of choice?

In my mind the axiom of choice is about being able to choose one element from each subset of some universal set $X$. I dont see the connection to the Axiom of choice in the part of the proof below. Let $u(x): \mathbb{R}^2 \rightarrow \mathbb{R}$…
stollenm
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Can one explain when the Axiom of Choice is needed in simple terms?

Consider the sequence of sets $(A_1,A_2,A_3,\dots)$. Assume for each $A_n$, there exists bijective mapping $f_n: \{1,2,3,4\} \rightarrow A_n$ for each $n$. If I want to construct the sequence $(f_1(2),f_2(2),f_3(2), f_4(2),\dots)$, do I need the…
Kun
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Does dense subset in a polish space contain a countable subset which is dense in the space?

Let $X$ be a separable complete metric space. Let $E$ be a dense subset of $X$ and fix $p\in X$. I want to 'choose' an element for each $\overline{B(p,1/n)}\cap E$ in ZF. ($n\in \mathbb{Z}^+$) Is it possible? As you can see in the link; Constructing…
Katlus
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Assuming a choice function for a set $S$

What I understand so far: If $S$ is any set then AC gives us a choice function, that is, $f: P(S)\setminus \{\varnothing \} \to S$ such that $f$ returns an element of $A \in P(S) \setminus \{\varnothing \}$. Assume we have a bijection $f: S \to…
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Choice of a real number using the axiom of choice

I wonder if one can define the axiom of choice by the assumption that is possible to choose an arbitrary real number? If we define real numbers as Cauchy limits of rational numbers, I assume that we must consider that as a set, since it is a case of…
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Naive understanding of choice axiom

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 finite set, there is exists some natural $n,…
mnaoumov
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Every set of pairwise disjoint sets has a choice function implies AC

Suppose that for every collection $A$ of non empty pairwise disjoint sets has a choice function. I need to prove that this implies the axiom of choice. Let $S$ be a collection of sets. For every $B$ in $S$, define $S_B = {\{}{(x,B)|x\in B}{\}}$. The…
Joshhh
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What does Axiom of Choice mean

So this a fundamental assumption in mathematics. Can someone explain informally what it actually is please. My guess is that its when we say in proofs that "Let $x \in X$". But I am not sure.
snowman
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Axiom of countable choice and injective functions

The axiom of countable choice states that for every countable collection $\{S_n\}_{n \in \mathbb{N}}$ of non empty sets there is a function $f : \mathbb{N} \to \bigcup_{n \in \mathbb{N}} S_n$ such that $f(n) \in S_n$ for every $n \in…
effezeta
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