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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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 ...
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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 ...
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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', ...
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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 ...
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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 ...
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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 ...
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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 ...
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Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ ...
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Over ZF, does “every Hilbert space have a basis” imply AC?

I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert ...
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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?
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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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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 ...
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Is Banach-Alaoglu equivalent to AC?

The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem. As I have ...
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A few questions about intuitionistic mathematics

I have to write a paper on Intuitionism for my Philosophy of Science class and I'm struggling with a few concepts I have encountered in my self-study. The (intuitive) characterization of valid ...
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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, ...
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Axiom of choice - to use or not to use

I was wondering if there are examples of results in mathematics that were first proven using axiom of choice and later someone found a proof of the result without using the axiom of choice.
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W. Mückenheim claims a severe inconsistency of transfinite set theory; true? [closed]

The abstract for a paper on arxiv.org (http://arxiv.org/pdf/math/0408089v3.pdf) reads (with my emphasis): "Transfinite set theory including the axiom of choice supplies the following basic theorems: ...
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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 $z_n \to a$ ...
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Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?

I know the usual proof of the existence of an algebraic closure for any field using Zorn's Lemma. The answer to this previous question makes it clear that in general, some nonconstructive axiom (not ...
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Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?

I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of ...
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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 ...
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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 ...
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Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$?

Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the ...
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Implication and Interpretation of Banach Tarski

As I understand, the Banach-Tarski paradox says a ball in 3-space may be decomposed into finitely many pieces and reassembled into two balls each of the same size as the original. Despite being called ...
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The Use of the Axiom of Choice in an Elementary Proof

I wanted to give some of the new undergrad analysis students the following problem: given the real numbers (with the standard topology, as they'd expect) one cannot have an uncountable set such that ...
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Defining cardinality in the absence of choice

Under ZFC we can define cardinality $|A|$ for any set $A$ as $$ |A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}. $$ This is because the axiom of choice allows any ...
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Proving “every set can be totally ordered” without using Axiom of Choice

It is known that the statement "every set can be totally ordered" is strictly weaker than Axiom of choice. How does one go about proving without using AC?
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For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$

Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally: $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$ If this statement is true, ...
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1answer
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Vector Spaces and AC

I know that the proof that every vector space has a basis uses the Axiom of Choice, or Zorn's Lemma. If we consider an axiom system without the Axiom of Choice, are there vector spaces that provably ...
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1answer
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Infinite prisoners with hats — is choice really needed?

The problem is this (recently asked about here): A countably infinite number of prisoners, each with an unknown and randomly assigned red or blue hat line up single file line. Each prisoner faces ...
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Importance of Axiom of Choice

First a quick question regarding the definition of the axiom of choice. Do the sets have to be mutually disjoint nonempty sets or just non-empty? One source states: "For any set X of nonempty sets, ...
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About a paper of Zermelo

This about the famous article Zermelo, E., Beweis, daß jede Menge wohlgeordnet werden kann, Math. Ann. 59 (4), 514–516 (1904), available here. Edit: Springer link to the ...
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Where do we need the axiom of choice in Riemannian geometry?

A friend of mine is a differential geometer, and keeps insisting that he doesn't need the axiom of choice for the things he does. I'm fairly certain that's not true, though I haven't dug into the ...
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1answer
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How to formulate continuum hypothesis without the axiom of choice?

Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers : 1) The definition of $\aleph_1$ makes sense even without choice as $\aleph_1$ is an ordinal number ...
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The cardinality of a countable union of countable sets, without the axiom of choice

One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have ...
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1answer
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Infinite Set is Disjoint Union of Two Infinite Sets

A finite set is a set such that there exists a bijection from it to some finite ordinal. An infinite set is a set that is not finite. In ZF, can you prove that every infinite set is the union of two ...
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A question about a two player game and axiom of choice

Suppose two players 1 and 2 play the following game: Player 1 starts by playing the set of reals $\mathbb{R}$. Player 2 plays an uncountable subset $Y_1$ of $\mathbb{R}$. Then player 1 plays an ...
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1answer
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Cardinality of a locally compact Hausdorff space without isolated points

I am interested in the following result: Theorem. A locally compact Hausdorff topological space $X$ without isolated points has at least cardinality $\mathfrak{c}$. To prove it, one can find two ...
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Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF

Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$. Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< ...
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Do we know that we can't define a well-ordering of the reals?

Folklore has it that it is impossible to define a well-ordering of the reals explicitly. There exist pointwise definable models of ZFC where every set is definable without parameters: it is the ...
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Axiom of Choice Examples

In the wikipedia article, two examples are given which use/ do not use the axiom of choice. They are: Given an infinite pair of socks, one needs AC to pick one sock out of each pair. Given an ...
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How far is it true that statements dependent on Axiom of Choice are not constructive.

Axiom of Choice is often used in mathematics to construct various objects, such as basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, unmeasurable subset of $\mathbb{R}$, or a non-principal ...
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Are sets constructed using only ZF measurable using ZFC?

Suppose $S$ is a subset of $\mathbb{R}$ which can be defined without using the axiom of choice, i.e. which can be proved to exist using only the axioms of ZF. Does it follow that $S$ is measurable? ...
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What is the “opposite” of the Axiom of Choice?

One might think that, trivially, the "opposite" of AC is $\neg$AC. However, thinking about it differently, I'm not sure this is intuitively the case. AC says that every set has a choice function. ...
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Do you need the Axiom of Choice to assert that every real vector space has a norm?

Math people: This question is 95% answered (the first answer) at Does every $\mathbb{R},\mathbb{C}$ vector space have a norm? and Vector Spaces and AC . The questions, answers, and links found there ...
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What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…

... $\aleph_1$ is the immediate successor of $\aleph_0$? I was reading the wiki article on $\aleph_1$ where a distinction is made between the two. If there's isn't a cardinal between $\aleph_1$ and ...
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Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set?

The existence of subsets of the real line which are not Lebesgue measurable can be argued using the Axiom of Choice. For example, define an equivalence relation on $[0, 1]$ by $a \thicksim b$ if and ...
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Prove that every set with more than one element has a permutation without fixed points

I cannot prove this statement so need help. This problem is one of exercises right after the chapter about Hausdorff's maximal principle and Zorn's Lemma. Thus, you cannot use the concept of cardinal ...
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Axiom of Choice and finite sets

So I am relatively familiar with the Axiom of Choice and a few of its equivalent forms (Zorn's Lemma, Surjective implies right invertible, etc.) but I have never actually taken a set theory course. I ...
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Axiom of Choice: Can someone explain the fallacy in this reasoning?

Not a "set theory" guru (apologies if my terms are imprecise), but I have heard that it is an elementary result that the set of rational numbers has a measure of zero - intuitively meaning that the ...