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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Mathematicians usually don't write "by the axiom of choice" when they prove some propositions even if they use the axiom of choice. But they ...

Mathematicians usually don't write "by the axiom of choice" when they prove some propositions even if they use the axiom of choice. But they always emphasize the fact they use the axiom of choice when they prove Zorn's Lemma. Why?
tchappy ha
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Axiom of choice and proving mathematical results

In many mathematical proofs dealing with mathematical objects, we begin by saying something along the lines of: Let $X$ be a such and such mathematical object. Let $y \in X$. And then we proceed to exploit the properties of $X$ to prove what we want…
zyx123
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Axiom of choice and function with empty codomain

I'm having a little problem here, namely if the axiom of choice (Wikipedia) is $$\forall X \left[ \emptyset \notin X \implies \exists f: X \to \bigcup X \quad \forall A \in X \, ( f(A) \in A ) \right]$$ and I choose the nonempty $X=\{\emptyset\}\neq…
Nikolaj-K
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Axiom of Choice and Completeness of the Reals

I have a question regarding the Completeness axiom of R. Does It have a relation with the axiom of choice? I mean, does the axiom of choice implies the Completeness axiom? Or are they independent?
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Countable chain covering a totally ordered set.

I wonder if this statement holds. I think it doesn't (if the cardinal of T is very large), but have no idea how to find a counter example. Let $T$ be a totally ordered set, does there exists a sequence $(t_n)_{n \in \omega}$ such that $ \forall t…
TuTor
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choosing $\lambda$ different elements inside a set of cardinality $\lambda^+$

my question is the following: I have a set $X$ $\lambda^+$-ordered (for $\lambda>\omega$) and I want to "extract" a subset $Y$ $\lambda$-ordered. Do I need AC to choose $\lambda$ different elements of $X$?
user647684
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Do we need AC to apply this proof of $U\subseteq \bigcup \mathscr S$?

To provide some context, I was trying to prove $(X, d)$ separable $\implies$ X second countable, specifically without using any form of choice. Let $U \subseteq X$ open and $D$ our countable dense subset, and consider $\mathscr B := \{B_q(d) \mid…
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Two parts of the axiom of choice

Let I be a set, and for each $\alpha \in I$, let $X_\alpha$ be a non-empty set. Then $\prod_{\alpha \in I}X_\alpha$ is also non-empty. There is (a) the possibly-infinite set $I$ and also (b) all the possibly-infinite sets $X_\alpha$. If one of…
Hatshepsut
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Zorn's Lemma Proof: Definition of Tower, Union Requirement

I was going through multiple proofs of Zorn's Lemma (Halmos', in Naive Set Theory, being amongst them), but I get stuck on the definition of towers, particularly the "union requirement" (requirement III in Halmos' proof and Proof Wiki). Say…
steve
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Existence of injection or surjection between two given sets

I'd like to know why, given two sets $X$ and $Y$, there always exists an injection from $X$ to $Y$ or a surjection from $X$ to $Y$. In other words, why we can always compare the cardinality of two sets. I appreciate any help!
Jiu
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What's wrong with my argument of choice function from infinite family of nonempty sets?

Let $(A_i \mid i \in I)$ be a family of non-empty set. Then $A_i \neq \emptyset$ for all $i \in I$, thus $\exists a_i \in A_i$ for all $i \in I$. We define $f:I \to \bigcup A_i$ by $f(i)=a_i$. From reading other posts in MSE, I'm sure that my…
Akira
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Proving the axiom of choice in propositions as types

The axiom of choice can be stated as follows for a relation $R$ between terms of type $A$ and type $B$: $$(\forall x : A.\exists y: B.R\; x\; y) \rightarrow \exists f:A\rightarrow B.\forall x : A.R\; x\; (f\; x)$$ In proposition as…
Eben Kadile
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Systematic review of the uses of the axiom of choice

Is there some systematic study of which mathematical results requires the axiom of choice? A book or a review article. I don't mean only results without which some areas of mathematics would be hard to work with (for example, that $\mathbb{R}$ is…
user450847
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Does the Axiom of choice allow the existence of a choice set before the end of its enunciation?

We can state the Axiom of Choice as follows: 'If A is a family of nonempty sets, then there is a function f with domain A such that f(a) ∈ a for every a ∈ A. Such a function f is called a choice function for A'. However, if A may be any family of…
Egli
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On axiom of choice

I have found in a divulgative book a claim which roughly says that if you construct a set with the axiom of choice, and you can describe it's elements with a proposition "p(x)", then you can avoid the axiom of choice in it's construction. Can…
HeMan
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