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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Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three classical consequences of the Baire category theorem in…
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Can you explain the "Axiom of choice" in simple terms?

As I'm sure many of you do, I read the XKCD webcomic regularly. The most recent one involves a joke about the Axiom of Choice, which I didn't get. I went to Wikipedia to see what the Axiom of Choice is, but as often happens with things like this,…
Tyler
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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 question is: Has anyone constructed a well ordering on…
Seamus
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What are the Axiom of Choice and Axiom of Determinacy?

Would someone please explain: What does the Axiom of Choice mean, intuitively? What does the Axiom of Determinancy mean, intuitively, and how does it contradict the Axiom of Choice? as simple words as possible? From what I've gathered from the…
user541686
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Predicting Real Numbers

Here is an astounding riddle that at first seems impossible to solve. I'm certain the axiom of choice is required in any solution, and I have an outline of one possible solution, but would like to see how others might think about it. $100$ rooms…
Jared
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Why can't you pick socks using coin flips?

I'm teaching myself axiomatic set theory and I'm having some trouble getting my head around the axiom of choice. I (think I) understand what the axiom says, but I don't get why it is so 'contentious', which probably means I haven't yet digested it…
MGA
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Continuity and the Axiom of Choice

In my introductory Analysis course, we learned two definitions of continuity. $(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ satisfies $f(z_n) \to f(a)$. $(2)$ A function $f:E \to…
John Gowers
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Does every set have a group structure?

I know that there is no vector space having precisely $6$ elements. Does every set have a group structure?
spohreis
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Foundation for analysis without axiom of choice?

Let's say I consider the Banach–Tarski paradox unacceptable, meaning that I would rather do all my mathematics without using the axiom of choice. As my foundation, I would presumably have to use ZF, ZF plus other axioms, or an approach in which sets…
user13618
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Explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$

Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? A related question is: Can we proved…
Seirios
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Axiom of choice and automorphisms of vector spaces

A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order $1$ or $2$." I think that the straightforward solution uses that a exponent two group is a vector space over $\operatorname{GF}(2)$,…
PseudoNeo
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Is the axiom of choice really all that important?

According to this book: The Axiom of Choice is the most controversial axiom in the entire history of mathematics. Yet it remains a crucial assumption not only in set theory but equally in modern algebra, analysis, mathematical logic, and topology…
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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 $f\colon \mathcal{P}^*(A)\rightarrow A$ such that for any subset $S$ of $A$,…
p Groups
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For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

How to prove the following conclusion: [For any infinite set $S$, there exists a bijection $f:S\to S \times S$] implies the Axiom of choice. Can you give a proof without the theory of ordinal numbers.
mathabc
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Why is the axiom of choice separated from the other axioms?

I don't know much about set theory or foundational mathematics, this question arose just out of curiosity. As far as I know, the widely accepted axioms of set theory is the Zermelo-Fraenkel axioms with the axiom of choice. But the last axiom seems…
bjorn
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