**126**

votes

**15**answers

24k views

### Why worry about the axiom of choice?

As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory. Yet I still see people fuss about whether or not theorem X depends on it, and I don't ...

**84**

votes

**2**answers

7k views

### Does every non-empty set admit a group structure (in ZF)?

It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary ...

**59**

votes

**16**answers

15k views

### Most 'unintuitive' application of the Axiom of Choice?

It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle, Tychonoff's theorem, and the fact that every vector space has a basis. Even ...

**42**

votes

**4**answers

10k views

### Non-Borel sets without axiom of choice

This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...

**40**

votes

**1**answer

2k views

### When does $A^A=2^A$ without the axiom of choice?

Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$
However without the axiom of choice this doesn't ...

**37**

votes

**4**answers

4k views

### Does the fact that this vector space is not isomorphic to its double-dual require choice?

Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their "...

**36**

votes

**1**answer

3k views

### Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice?

The Generalized Continuum Hypothesis can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, The Axiom of Choice, Theorem 9.1 p.133).
In fact, a relatively weak ...

**35**

votes

**6**answers

2k views

### Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The ...

**32**

votes

**6**answers

2k views

### Distinct well-orderings of the same set

An easy consequence of the Erdős-Dushnik-Miller theorem $\kappa\to(\kappa,\omega)^2$ is the following, that will denote $(*)$ (it appears as an exercise in Kunen's book, it was probably mentioned ...

**32**

votes

**1**answer

2k views

### Dual Schroeder-Bernstein theorem

This question was motivated by the comments to Dual of Zorn's Lemma?
Let's denote by the Dual Schroeder-Bernstein theorem (DSB) the statement
For any sets $A$ and $B$, if there are ...

**30**

votes

**2**answers

1k views

### On the difference between two concepts of even cardinalities: Is there a model of ZF set theory in which every infinite set can be split into pairs, but not every infinite set can be cut in half?

An interesting question has arisen over at this
math.stackexchange
question
about two concepts of even in the context of infinite
cardinalities, which are equivalent under the axiom of
choice, but ...

**29**

votes

**0**answers

1k views

### Concerning the various proofs from the axiom of choice that R^3 admits of surprising geometrical decompositions into circles, skew lines and so on: can we prove in any instance that there are no Borel such decompositions? Or that AC is required?

This question follows up on a comment I made on Joseph O'Rourke's
recent question, one of several questions here on mathoverflow
concerning surprising geometric partitions of space using the axiom
of ...

**28**

votes

**15**answers

2k views

### Objects which can't be defined without making choices but which end up independent of the choice

It happens a lot of times that when one defines a new object (ring, module, space, group, algebra, morphism, whatever) out of given data one first chooses some additional structure. And sometimes (...

**28**

votes

**6**answers

4k views

### Why can't proofs have infinitely many steps?

I recently saw the proof of the finite axiom of choice from the ZF axioms. The basic idea of the proof is as follows (I'll cover the case where we're choosing from three sets, but the general idea is ...

**25**

votes

**2**answers

790 views

### How much choice is needed to show that formally real fields can be ordered?

Background: a field is formally real if -1 is not a sum of squares of elements in that field. An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you ...

**24**

votes

**4**answers

3k views

### Are all sets totally ordered ?

The question is the title.
Working in ZF, is it true that: for every nonempty set X, there exists a total order on X ?
If it is false, do we have an example of a nonempty set that has no total ...

**23**

votes

**2**answers

2k views

### Hahn's Embedding Theorem and the oldest open question in set theory

Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his paper Über die nichtarchimedischen Grössensysteme (Sitzungsberichte der ...

**22**

votes

**3**answers

2k views

### Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?

The existence and uniqueness of algebraic closures is generally proven using Zorn's lemma. A quick Google search leads to a 1992 paper of Banaschewski, which I don't have access to, asserting that ...

**22**

votes

**1**answer

2k views

### If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?

Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if:
Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
If $\sum\alpha_i b_i = 0$, where $\...

**22**

votes

**0**answers

459 views

### Can one divide by the cardinal of an amorphous set?

This question arose in a discussion with Peter Doyle.
It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \{1,\ldots,k\} \simeq Y \times \{1,\ldots,k\}$ then $X ...

**22**

votes

**0**answers

904 views

### Supercompact and Reinhardt cardinals without choice

A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:
Definition. A cardinal $\kappa$ is supercompact if for all ordinals $\...

**21**

votes

**2**answers

2k views

### Does Arzelà-Ascoli require choice?

Inspired by a recent Math.SE question entitled Where do we need the axiom of choice in Riemannian geometry?, I was thinking of the Arzelà--Ascoli theorem. Let's state a very simple version:
...

**21**

votes

**2**answers

1k views

### Axiom of Choice: Ultrafilter vs. Vitali set

It is well known that from a free (non-principal) ultrafilter on $\omega$ one can define a non-measurable set of reals. The older example of a non-measurable set is the Vitali set,
a set of ...

**21**

votes

**3**answers

2k views

### Probabilities in a riddle involving axiom of choice

The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question):
The Riddle:
We assume ...

**21**

votes

**1**answer

653 views

### Is there a model of ZF set theory with a set that does not inject into the cardinals?

Question. Is there is a model of ZF set theory with a set $X$ that does not inject into the cardinals?
I use the term "cardinal" here in the ZF sense, so they are not necessarily well-orderable.
To ...

**20**

votes

**1**answer

1k views

### Splitting infinite sets

There are two questions here, an explicit one, and another (more vague) one that motivates it:
I am pretty certain the following should have a negative answer, but at the moment I'm not seeing how to ...

**20**

votes

**0**answers

1k views

### Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice

While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then ...

**19**

votes

**3**answers

2k views

### Half Cantor-Bernstein Without Choice

I had a discussion with one of my teachers the other day, which boiled to the following question:
Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\...

**19**

votes

**4**answers

2k views

### What are the known implications of “There exists a Reinhardt cardinal” in the theory “ZF + j”?

This is, alas, in large part a series of questions on unpublished work of Hugh Woodin; it's also quite frivolous if Reinhardt cardinals turn out inconsistent.
Definitions:
Call $\kappa$ an $I-1(...

**19**

votes

**0**answers

371 views

### Are the reals really a fraction field?

In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$.
A crucial point in the ...

**19**

votes

**2**answers

2k views

### Can a Vitali set be Lebesgue measurable? (ZF)

Here is the definition of Lebesgue measure.
The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In ...

**18**

votes

**1**answer

991 views

### Linear Algebra without Choice

We consider the field of "usual" linear algebra.
Q. Which aspects of it can be carried out without the Axiom of Choice?
Q. Do interesting "exotic" phenomena appear in presence of (some instance of) ...

**18**

votes

**1**answer

324 views

### Linear maps between arbitrarily chosen vectors of vector spaces $V$ and $W$

I recently came across this question:
Is the axiom of choice needed to prove the following statement:
Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. ...

**18**

votes

**1**answer

1k views

### The Continuum Hypothesis and Countable Unions

I recently edited an answer of mine on math.SE which discussed the implication of the two assertions:
$AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and
$CH$ which says that if $A\subseteq 2^{\omega}$ ...

**17**

votes

**2**answers

1k views

### What is a Choice Principle, really?

This question is quite soft, and I apologize in advance if it borderline off-topic.
When working in theories between ZF and ZFC the term "choice principle" is heard quite often. For example:
$\quad$ ...

**17**

votes

**4**answers

2k views

### Nilradicals without Zorn's lemma

It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.
Every proof I found (e.g. in the classical "Commutative Algebra" by ...

**17**

votes

**6**answers

1k views

### Unique Existence and the Axiom of Choice

The axiom of choice states that arbitrary products of nonempty sets are nonempty.
Clearly, we only need the axiom of choice to show the non-emptiness of the product if
there are infinitely many ...

**17**

votes

**1**answer

769 views

### Does ZF prove that topological groups are completely regular?

Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$.
Assume $\{\mathbf{e}\}$ is closed in $\langle G,\...

**17**

votes

**0**answers

597 views

### Antichains of Cardinals in ZF Without Choice…

With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what ...

**16**

votes

**1**answer

1k views

### Axiom of choice and bases of vector spaces over a fixed field

Let $k$ be a field. In 1984 Andreas Blass proved that the axiom "for every extension $K|k$, every vector space over $K$ has a basis" implies the axiom of choice. He also raised the question
Does ...

**16**

votes

**2**answers

757 views

### Haar measures in Solovay's model

Haar measure is a measure on locally compact abelian groups which is invariants to translations. For example, the Lebesgue measure on the reals is such measure.
It can be shown without the use of the ...

**16**

votes

**1**answer

583 views

### Categorifications of Zorn's lemma

I'm wondering about categorifications of Zorn's lemma along the following lines.
Lemma: if $\mathbf{C}$ is a small category in which every directed diagram of monomorphisms has a cocone of ...

**15**

votes

**11**answers

4k views

### Does the Axiom of Choice (or any other “optional” set theory axiom) have real-world consequences? [closed]

Or another way to put it: Could the axiom of choice, or any other set-theoretic axiom/formulation which we normally think of as undecidable, be somehow empirically testable? If you have a particular ...

**15**

votes

**5**answers

2k views

### Compactness of the Hilbert cube without the Axiom of Choice

I am just curious: is there a published proof of the compactness of the Hilbert cube that does not use the Axiom of Choice, or is it well known?

**15**

votes

**4**answers

2k views

### Is it possible to formulate the axiom of choice as the existence of a survival strategy?

Consider the following situation:
There is an infinite set $G$ of giraffes.
A lion comes and announces a set $C$ of all possible colours and an infinite cardinal $\kappa$.
The hungry lion ...

**15**

votes

**4**answers

1k views

### What sort of large cardinal can $\aleph_1$ be without the axiom of choice?

Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal.
...

**15**

votes

**1**answer

1k views

### How much of GCH do we need to guarantee well-ordering of continuum?

It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, ...

**15**

votes

**1**answer

873 views

### Sizes of bases of vector spaces without the axiom of choice

Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two ...

**15**

votes

**1**answer

2k views

### Countable Unions And The Axiom Of Countable Choice

Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of ...

**15**

votes

**1**answer

852 views

### Is Dependent Choice equivalent to the statement that every PID is factorial?

In this question, it was asked if AC is needed in the proof of the well-known fact that every principal ideal domain is factorial. As KConrad and Joel David Hamkins have pointed out, only DC, the ...