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22
votes
3answers
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 ...
15
votes
1answer
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: ...
1
vote
4answers
888 views

Choice Function on the Powerset of the Reals

I'm not sure if this question is appropriate for mathoverflow, but I can't help but think that other people have wondered about it as well. When anyone first learns about the axiom of choice, the ...
3
votes
1answer
356 views

Existence of enough projectives in the category of sets

I am talking about the principle that says that every set is the image of a projective set. For every set $x$ there is a surjection $f:y \twoheadrightarrow x$, such that for any set $u$ and function ...
3
votes
2answers
139 views

Global or Relativised Dependent Choices

I am talking about the principle that is to DC what the global choice is to the usual axiom of choice. Global choice involves existential quantification over classes, but global DC can be stated as a ...
6
votes
4answers
3k views

Does constructing non-measurable sets require the axiom of choice?

The classic example of a non-measurable set is described by wikipedia. However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} ...
31
votes
6answers
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
1answer
813 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
1answer
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
2answers
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 / ...
22
votes
4answers
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
6answers
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
3answers
799 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
2answers
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
2answers
857 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
4answers
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
4answers
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 ...
15
votes
11answers
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 ...
15
votes
4answers
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
2answers
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
2answers
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
3answers
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
8answers
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 ...
106
votes
15answers
19k 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
0answers
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
1answer
967 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
1answer
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
5answers
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
2answers
761 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
2answers
748 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
15answers
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
2answers
488 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
5answers
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
2answers
645 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 ...
65
votes
3answers
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
4answers
757 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 ...
22
votes
2answers
691 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
6answers
993 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
5answers
911 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.