The axiom-of-choice tag has no wiki summary.

**31**

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 ...

**14**

votes

**1**answer

811 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 ...

**31**

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 ...

**6**

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 / ...

**21**

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 ...

**25**

votes

**6**answers

3k 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 ...

**10**

votes

**3**answers

794 views

### Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice.

It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the ...

**9**

votes

**2**answers

1k views

### Proving Independence of Axioms by Exhibiting Models Which Don't Satisfy Our Intuition

I recently saw the proof of the independence of ZF (with allowance for multiple empty sets) and AC. The proof constructed the model based on a set theory generated by infinitely many empty sets and ...

**7**

votes

**2**answers

855 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?

**34**

votes

**4**answers

8k 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 ...

**10**

votes

**4**answers

3k views

### Finite axiom of choice: how do you prove it from just ZF?

The axiom of choice asserts the existence of a choice function for any family of sets F. Suppose, however, that F is finite, or even that F just has one set. Then how do we prove the existence of a ...

**14**

votes

**11**answers

3k 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 ...

**14**

votes

**4**answers

1k 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 ...

**3**

votes

**2**answers

1k views

### Axiom of Computable Choice versus Axiom of Choice

What would be the consequence of requiring that any choice function be computable; i.e. using as the foundational basis ZF + ACC? Does it make a difference if we admit definable functions?
I guess I ...

**11**

votes

**2**answers

2k views

### AC in group isomorphism between R and R^2

Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of ...

**6**

votes

**3**answers

3k views

### Cardinality: Why is there no “ℵ½”?

A wikipedia page/paragraph on ℵ₁ states:
"The definition of ℵ₁ implies (in
ZF, Zermelo-Fraenkel set theory
without the axiom of choice) that no
cardinal number is between ℵ₀ and
ℵ₁."
"If the axiom ...

**8**

votes

**8**answers

2k views

### Choice vs. countable choice

This question arose after reading the answers (and the comments to the answers) to Why worry about the axiom of choice?.
First things first. In my intuitive conception of the hierarchy of sets, the ...

**95**

votes

**15**answers

18k 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 ...

**6**

votes

**0**answers

1k views

### Explicit element of $(\ell^{\infty})^* - \ell^1$? [duplicate]

Possible Duplicate:
What’s an example of a space that needs the Hahn-Banach Theorem?
It is well known that the dual of $\ell^{\infty}$ properly contains $\ell^1$ (over $\mathbb{N}$, ...

**8**

votes

**1**answer

958 views

### Why does this sum depend on the Axiom of Choice?

On page 168 of Mathematical Fallacies and Paradoxes, it states that the fact that the series
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $
has a sum depends on the Axiom of Choice. Where ...

**2**

votes

**1**answer

2k views

### What is a universal function?

This question stems from Dick Lipton's recent blog post on the Axiom of Choice. I asked there but got no takers. I promise I'm not an inept Googler, but I couldn't find a satisfactory answer. I ...

**8**

votes

**5**answers

2k 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 ...

**3**

votes

**2**answers

756 views

### Is it still impossible to partition the plane into Jordan curves without choice?

It is an easy exercise to show that the Euclidean plane cannot be partitioned into round circles (note however that it is possible to do so for $\mathbb{R}^3$). It seems almost obvious that it is not ...

**12**

votes

**2**answers

744 views

### Ultrafilters vs Well-orderings

This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also ...

**47**

votes

**15**answers

12k 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 ...

**6**

votes

**2**answers

481 views

### Choice function for Borel sets?

Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. ...

**2**

votes

**5**answers

1k views

### Decidability of the Axiom of Choice

Everything that I read regarding Set Theory states that the Axiom of Choice is independent and undecidable within the context of Zermelo-Frankel Set Theory. My question is this: Is there any ...

**1**

vote

**2**answers

640 views

### Axiom of Choice and Order Types

A beginner's question:
We know: "Since order-equivalence is an equivalence relation, it partitions the class of all sets into equivalence classes." (from Wikipedia)
This holds since every set can be ...

**66**

votes

**3**answers

6k 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 ...

**2**

votes

**4**answers

753 views

### Set theories that do require the existence of urelements?

I am looking for an axiomatic set theory that not only admits the existence of urelements/atoms (via two-sortedness or an additional unary predicate) but requires it, e.g. by an axiom like "for each ...

**20**

votes

**2**answers

670 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 ...

**2**

votes

**6**answers

990 views

### Splitting lemma under assumption of the axiom of choice

The splitting lemma says:
Given a short exact sequence with maps $q$ and $r$:
$0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$
then the following are ...

**11**

votes

**5**answers

904 views

### Where are some interesting places where the axiom of choice crops up in category theory?

The two that come to mind are splitting epics in Set and taking the Skel of a category. Surely there are lots of other interesting (and maybe upsetting) places where this comes up.