**125**

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 ...

**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 ...

**14**

votes

**2**answers

1k views

### Does ZFC prove the universe is linearly orderable?

It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example ...

**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 ...

**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 ...

**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 ...

**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 ...

**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 ...

**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 ...

**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 "...

**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 ...

**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 ...

**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 $\...

**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:
$\quad$ ...

**15**

votes

**1**answer

872 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

**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.
...

**10**

votes

**2**answers

507 views

### Is sigma-additivity of Lebesgue measure deducible from ZF?

Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)?
Note 1. ...

**6**

votes

**1**answer

192 views

### Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?

A result commonly, and probably erroneously, attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on ...

**12**

votes

**5**answers

902 views

### Does k(X) have a k-basis for every set X, without AC?

This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.
For any field $k$, the field $k(x)$ of rational functions in one variable has an ...

**6**

votes

**0**answers

258 views

### Does Sageev's result need an inaccessible?

In 1981, building on work by Ellentuck in 1974, Sageev showed ("A model of ZF + there exists an inaccessible, in which the Dedekind cardinals constitute a natural non-standard model of arithmetic," ...

**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:
...

**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 ...

**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 ...

**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 ...

**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 ...

**9**

votes

**5**answers

3k views

### Subset of the plane that intersects every line exactly twice

In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to ...

**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 ...

**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 ...

**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 ...

**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 $A\subseteq 2^{\omega}$ ...

**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(...

**14**

votes

**0**answers

389 views

### The axiom $I_0$ in the absence of $AC$

It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$
then $\lambda$ has countable cofinality (and in ...

**13**

votes

**4**answers

1k views

### Forcing over models without the axiom of choice

In the vast majority of papers forcing is always developed over ZFC.
Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain ...

**11**

votes

**1**answer

3k views

### Are there any non-linear solutions of Cauchy's equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice?

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$
It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in ...

**8**

votes

**0**answers

571 views

### Full conditional probabilities and versions of AC?

A probability is a finitely additive measure on a boolean algebra with total measure $1$.
A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean ...

**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?

**10**

votes

**1**answer

880 views

### Can an infinite number of mathematicians guess the number in a box with only one error?

In this question the following observation was made:
Consider a sequence of boxes numbered 0, 1, ... each containing one real number. The real number cannot be seen unless the box is opened.
Define ...

**7**

votes

**3**answers

1k views

### Existence of prime ideals and Axiom of Choice.

One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple ...

**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 ...

**14**

votes

**1**answer

473 views

### Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?

The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the ...

**12**

votes

**1**answer

503 views

### Without choice, can every homomorphism from a profinite group to a finite group be continuous?

In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the ...

**7**

votes

**2**answers

898 views

### Can iterating countable unions give every set? (ZF)

Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under at-most-countable unions?

**7**

votes

**2**answers

1k views

### Compact Hausdorff spaces without isolated points in ZF

S is uncountable := |$\mathbb{N}$| < |S|
S is noncountable := |S| $\not\leq |\mathbb{N}|$
(X,$T$) is a nice space := (X,$T$) is a compact Hausdorff space without isolated points
Does [ ZF / ...

**6**

votes

**1**answer

461 views

### Strictly order preserving maps into the integers

If $P$ and $P'$ are partial orders, a strictly order preserving map from $P$ to $P'$ is an $f:P\to P'$ satisfying that $x\lt y$ implies $f(x)\lt f(y)$ for all $x,y\in P$.
An interval in $P$ is a set ...

**4**

votes

**5**answers

1k views

### What axioms are stronger than the Axiom of choice?

What other axioms in set theory are stronger than AC ? I mean what are those axioms that will imply AC ?

**12**

votes

**1**answer

547 views

### Can there be a global linear ordering of the universe without a global well-ordering of the universe?

This question arose in the answers to Asaf Karagila's
question Does ZFC prove that the universe is linearly orderable?. The answer there was that one can have a ZFC model with no global linear ...

**10**

votes

**4**answers

730 views

### On surjections, idempotence and axiom of choice

The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF?
(I couldn't find it in the Consequences site so far.)
If $A$ is an infinite set such that $A$ ...

**9**

votes

**1**answer

321 views

### Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...

**9**

votes

**1**answer

309 views

### Cardinals without choice: interpolation (reference wanted)

Is there a published reference for this ZF theorem?
Let $m,n\in\mathbb{N}$. If $a_1,\dots,a_m$ and $b_1,\dots,b_n$ are cardinals such that $a_i\le b_j$ for all $i$ and $j$, then there is a cardinal $...