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 ...
69
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4answers
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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 ...
59
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6answers
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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', ...
49
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2answers
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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 ...
47
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3answers
15k views

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 ...
39
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2answers
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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 ...
33
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2answers
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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 $L^r$ ...
28
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4answers
1k views

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?
26
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5answers
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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.
25
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2answers
661 views

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 ...
23
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4answers
952 views

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 ...
22
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5answers
781 views

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.
20
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4answers
689 views

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 ...
18
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2answers
1k views

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 ...
18
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2answers
971 views

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 $z_n \to a$ ...
18
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4answers
941 views

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 ...
18
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3answers
734 views

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 ...
17
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6answers
650 views

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 ...
17
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3answers
901 views

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 ...
16
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2answers
822 views

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 ...
16
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1answer
928 views

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?
15
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5answers
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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, ...
15
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3answers
594 views

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, ...
15
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4answers
906 views

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 ...
15
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1answer
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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 ...
15
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1answer
305 views

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 ...
15
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1answer
942 views

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 ...
15
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2answers
378 views

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 ...
14
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2answers
630 views

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}< ...
14
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2answers
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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 ...
14
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1answer
573 views

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 ...
14
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2answers
155 views

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 ...
14
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1answer
317 views

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 ...
13
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6answers
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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, ...
13
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3answers
570 views

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 ...
13
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1answer
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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 ...
13
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2answers
257 views

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 ...
12
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4answers
543 views

Number Theory in a Choice-less World

I was reading this article on the axiom of choice (AC) and it mentions that a growing number of people are moving into school of thought that considers AC unacceptable due to its lack of constructive ...
12
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3answers
231 views

Axiom of Choice and Determinacy

In my set theory course we have been talking about the axiom of determinacy. One of the first things we showed was that $AD$ and $AC$ are incompatible. We later showed that $ZF+AD$ implies the ...
12
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2answers
340 views

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 ...
12
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2answers
642 views

Transfinite Induction and the Axiom of Choice

My question is essentially this: Why does the principle of transfinite induction not suffice to show the axiom of choice when the sets to be chosen from are indexed by a well ordered set? I have read ...
12
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1answer
260 views

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 ...
12
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1answer
407 views

2 questions about “$\mathbb R$ has a Hamel basis over $\mathbb Q$”

I would like to know if there are "interesting" equivalences to the statement in the title. I am not interested in more general statements, like "every vector space has a Hamel basis" or "every vector ...
12
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1answer
207 views

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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0answers
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Is Dover publishing Moore's book on the Axiom of Choice? [closed]

Dover is publishing a paperback edition of Gregory H. Moore's Zermelo's Axiom of Choice: Its Origins, Development, and Influence. It's supposed to come out March 20th and is available for pre-order at ...
11
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1answer
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Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people: It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. ...
11
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2answers
867 views

Uncountable subset with uncountable complement, without the Axiom of Choice

Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is ...
11
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1answer
468 views

There's non-Aleph transfinite cardinals without the axiom of choice?

I can't find anything on this anywhere. The book I'm largely using at the moment is based around ZFC, so it makes no mention of anything other than the Aleph numbers, but according to Wikipedia on the ...
11
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2answers
714 views

Equivalent statements of the Axiom of Choice

As a little project for myself this winter break, I'm trying to go through as much of Enderton's Elements of Set Theory as I can. I hit a snag trying to show two forms of the Axiom of Choice are ...
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177 views

The relationship of ${\frak m+m=m}$ to AC

Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.) Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$? If not, what is the ...