**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 ...

**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 ...

**6**

votes

**1**answer

453 views

### Symmetric extensions and class forcing

Suppose $V\models ZFC$ and $P\in V$ is a poset of forcing conditions.
It is a basic theorem in forcing that $V[G]\models ZFC$ for any generic extension by a $V$-generic filter $G$.
It is also known ...

**4**

votes

**1**answer

310 views

### Does ZF prove that proximity spaces are completely regular?

(This is based on my earlier question, but I think this one would be easier to answer.)
Let $\langle X,\mathbf{\delta} \hspace{.01 in} \rangle$ be a separated proximity space, and let $\cal{T}\hspace{...

**9**

votes

**3**answers

747 views

### On a weak choice principle

[PLEASE SEE EDITS AT BOTTOM OF QUESTION]
Consider the following set-theoretic axiom:
For each set $X$ there exists a set-indexed collection $\{C_i \to X\}_{i\in I_X}$ of surjections such that for ...

**3**

votes

**1**answer

2k views

### Axiom of choice and non measurable set

We know that existence of a Lebesgue non-measurable set is consistent with the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue non-measurable set imply the Axiom Of ...

**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,\...

**2**

votes

**1**answer

365 views

### Can a Vitali Set be constructed without AC?

For the purposes of this discussion, let a Vitali Set be any subset $V\subseteq{}[0,1)$ such that for $V_q:=\{x+q\;|\;x<1-q,\;x\in{}V\}\cup\{x+q-1\;|\;x\geq{}1-q,\;x\in{}V\}$ there is a countable ...

**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 ...

**7**

votes

**1**answer

343 views

### For models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$?

Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$.
Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if $...

**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 ...

**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 ...

**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 ...

**6**

votes

**5**answers

950 views

### Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?

In ZFC, every construction of a Lebesgue or Borel non-measurable set uses the axiom of choice. None of them that I've seen use choice to define a unique set, even though it's entirely possible to do ...

**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 ...

**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\...

**11**

votes

**2**answers

840 views

### Indecomposable vector spaces and the axiom of choice

It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B)
It is also known ...

**2**

votes

**0**answers

271 views

### algebraic dual and axiom of choice [duplicate]

If $K$ is a field, the dual of $K^{({\mathbb N})}$ is $K^{\mathbb N}$, and axiom of choice implies that the natural map from $K^{({\mathbb N})}$ to the dual of $K^{\mathbb N}$ is
far from being ...

**4**

votes

**3**answers

741 views

### Chevalley's valuation extension theorem and the axiom of choice

Hello,
Do we know if the axiom of choice is needed for Chevalley's valuation/place extension theorem (i.e. the theorem that states that for every valued field and a field extension, one can extend ...

**6**

votes

**3**answers

748 views

### Reference Request: Independence of the ultrafilter lemma from ZF

I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"):
The ultrafilter lemma is independent of ZF.
ZF + the ultrafilter ...

**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 ...

**4**

votes

**1**answer

528 views

### How much choice do we need for regularity of product of regular spaces ?

It is usually stated that the (possibly uncountable) product
of regular topological spaces is regular.
However the only proof that I know of this fact seems to use the full axiom of choice :
See ...

**2**

votes

**2**answers

795 views

### Countable Fields with No Countable Extension

Let $\mathscr{S}$ be the set of all countable subfields of $\mathbb{C}$. Clearly, $\mathscr{S}$ is a partially ordered set under inclusion, and if $K_1\subseteq K_2 \subseteq \cdots$ is an ascending ...

**1**

vote

**4**answers

837 views

### If a result is apparently provable with AC, is actually independent of ZF?

Given the number of results that are independent of ZF. It seems that once you've found a proof of a theorem that uses the axiom of choice, the odds are that it will be independent of ZF. So my ...

**8**

votes

**1**answer

627 views

### Axiom of Choice in a weaker system

Is it known whether or not there is a consistent system of logic where two or all of the axiom of choice, well-ordering principle, and Zorn's lemma have no (known) proof of equivalence?
I was ...

**8**

votes

**3**answers

3k views

### Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice?

Let X be an infinite set.
Is it possible to show the existence of a countably infinite subset of X without using the Axiom of Choice?

**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 ...

**2**

votes

**1**answer

738 views

### Is “second-countable implies separable” equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for ...

**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 ...

**7**

votes

**3**answers

742 views

### Construction of a maximal ideal

Hello,
Let R denote the ring of continuous functions defined on the real line, let I in R be the ideal consisting of functions with compact support. Obviously, I is not maximal, and by Zorn's Lemma ...

**2**

votes

**1**answer

416 views

### Nets and the Axiom of Choice

Suppose that $f : X \rightarrow Y$ is mapping between topological spaces that is not continuous at $x_0$. Then there is an open set $V$ in $Y$ containing $f(x_0)$ such that for any open set $U$ ...

**11**

votes

**1**answer

682 views

### Dynamical systems, minimal sets and the Axiom of Choice

Perhaps the most important application of the Axiom of Choice within the theory of dynamical systems (meaning here, compact Hausdorff spaces with a self-map) yields, within every dynamical system, the ...

**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 ...

**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 "...

**9**

votes

**2**answers

412 views

### Completeness of the club filter without AC

Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the ...

**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

**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:
...

**1**

vote

**4**answers

997 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

**1**answer

362 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

**2**answers

145 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

**4**answers

4k 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} /\...

**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 ...

**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 ...

**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 ...

**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 / ...

**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 ...

**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 ...

**10**

votes

**3**answers

860 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 ...

**11**

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

899 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?