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**1**answer

160 views

### Partial Universes and the Axioms of $ZF$ Set Theory Without Choice

In his Senior Thesis, Samuel Coskey answered the question of which axioms of $ZFC$ hold at each stage of the cumulative hierarchy. Here is the list of his results:
Axioms that always hold: ...

**1**

vote

**1**answer

122 views

### Without Choice: Are there filters of cardinality continuum?

Is it provable, in ZF (without Choice), that every filter can be extended to one of cardinality continuum?
The extended filter is not requested to be an ultrafilter.

**7**

votes

**0**answers

102 views

### Choice and the Baire property in non-separable complete metric spaces

It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has ...

**3**

votes

**1**answer

98 views

### What are the minimal requirements for the definable hyperreal field plus transfer?

It is interesting that to prove the transfer principle for the definable hyperreal field, one requires no more choice than for proving, for instance, the countable additivity of the Lebesgue measure. ...

**5**

votes

**2**answers

209 views

### Constructively, are all fibrations cloven?

A "cloven fibration" is a fibration for which we have an explicit choice of cartesian liftings; this is often phrased as, "We can pick a lifting without using the axiom of choice".
Firstly, I'm a bit ...

**3**

votes

**1**answer

124 views

### $2M = M$ and its subset

I have some question concerning arithmetic of cardinal in ZF.
Write $ X = Y$ if there is a bijection between them.
Let $M$ be a set such that $2M = M$.
Can I show, in ZF, that any infinite subset $X$ ...

**1**

vote

**1**answer

206 views

### Well-ordering of power set of $\omega$

Assuming initially the background set theory ZF, what is the exact status of the existence of a well-ordering on the power set of $\omega$? How much needs to be added to guarantee this?

**2**

votes

**1**answer

176 views

### Transfer with minimal choice

Let FUF postulate the existence of a Free UltraFilter on $\mathbb{N}$ and ACC the axiom of countable choice. Consider the superstructure on $\mathbb{R}$ and its inclusion in the bounded ultrapower. ...

**1**

vote

**1**answer

158 views

### Is there any infinite set which is Dedekind finite and weakly Dedekind finite?

Is there any infinite set which is Dedekind finite and weakly Dedekind finite?
If $X$ is a weakly Dedekind finite amorphous set, can we show that $\mathcal P(X)$, the power set of $X$, is also weakly ...

**14**

votes

**0**answers

391 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 ...

**14**

votes

**1**answer

480 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 ...

**23**

votes

**0**answers

468 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 ...

**11**

votes

**0**answers

225 views

### Is there a particular field that cannot be proven to have an algebraic closure in ZF?

Proofs that every field has a unique (up to isomorphism) algebraic closure use some form of the axiom of choice. For uniqueness this is provably necessary: there are models of ZF in which $\mathbb{Q}$ ...

**3**

votes

**2**answers

296 views

### Linear space with (Hamel) basis and the axiom of choice

It is true that the axiom of choice is equivalent to the statement that every linear space has a Hamel basis. There are some linear spaces which definitely don't need axiom of choice to possess (...

**2**

votes

**1**answer

192 views

### Existence of Spanning Tree implies Well Ordering Principle

Every connected graph has a spanning tree.
Every non-empty set can be well ordered.
Basically I am trying to show that statement 1 implies statement 2. What I tried is as following:
Let $X \ne \...

**6**

votes

**1**answer

217 views

### Does “$|{\cal P}_2(X)| = |X|$ for $X$ infinite” imply ${\sf (AC)}$?

This comes from a comment made by user bof in this thread.
Let $X$ be a set, define ${\cal P}_2(X) = \big\{\{a, b\}: a\neq b\in X\big\}$.
Consider the statement
${\sf (S)}$ If $X$ is an ...

**5**

votes

**1**answer

287 views

### How much choice does a linear or well-order on cardinals imply?

It is well-known that if the natural (partial) order on the class of cardinal numbers is a linear order, then it is in fact a well-order and the axiom of choice holds. I was, however, interested in ...

**12**

votes

**0**answers

286 views

### Is the universality of the surreal number line a weak global choice principle?

I'd like to consider the principle asserting that the surreal
number line is universal for all class linear orders, or in other
words, that every linear order (including proper-class-sized)
linear ...

**10**

votes

**1**answer

422 views

### What sort of large cardinal can continuum be?

I have stumbled across a related question asking which large cardinal properties can hold for $\aleph_1$. This question is probably also related, asking in what ways $\aleph_0$ is a "large" cardinal.
...

**4**

votes

**1**answer

314 views

### Is tree property inconsistent with Berkeley cardinals in the absence of Axiom of Choice?

On one hand due to Kunen's inconsistency theorem it is known that within $\sf ZF$, large cardinal axioms beyond Reinhardt cardinal are inconsistent with $\sf AC$.
Also some recent results of Bagaria, ...

**13**

votes

**1**answer

486 views

### Axiom of choice as zero dimensionality

In the paper Quantifiers and Sheaves by Lawvere, at the bottom of the second page, the author writes:
"... the condition that every epi splits, which geometrically we would call 0-dimensionality ...

**4**

votes

**2**answers

367 views

### Maximal chains and antichains of statements weaker than AC

First, I would like to say that I asked this question (a more general one actually) on math.stackexchange.com.
Consider the set $\varPhi$ of statements in the language of ZF that are weaker than AC (...

**3**

votes

**0**answers

95 views

### Does Łoś's theorem imply choice given a free ultrafilter?

In the paper "Łoś's theorem and the boolean prime ideal theorem imply axiom of choice" Howard has shown that Łoś's theorem and the boolean prime ideal theorem imply axiom of choice. At the end of the ...

**6**

votes

**1**answer

193 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 ...

**6**

votes

**0**answers

223 views

### Axiom of choice and a set in the plane that intersects every line in two points

In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects ...

**6**

votes

**1**answer

311 views

### Finding non convex functions satisfying a weak form of convexity, without the axiom of choice

If a real-valued function $f$ over reals satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand,...

**3**

votes

**0**answers

310 views

### How close to being well-orderable does this make my powerset?

Let's work in a set theory without assuming AC (for instance, but not necessarily, ZF). Fix a set $k$ satisfying $k\times k \simeq k$, and consider its powerset $X = 2^k$. I have a technical condition ...

**5**

votes

**1**answer

607 views

### Replacing Axiom of Choice with Axiom of Countable Choice

Many people find ACC more intuitive than AC ("Pick something from the first set, then something from the second set, then...) and it also doesn't lead to "controversial consequences" (See for eg: ...

**4**

votes

**0**answers

178 views

### On the Axiom of Choice for Conglomerates and Skeletons

Say that $\mathcal{X}$ is a conglomerate if $\mathcal{X} = \{X_i: i \in I\}$, where each $X_i$ and $I$ are classes. The Axiom of Choice for Conglomerates is the statement: Whenever $\mathcal{X}$ and $\...

**3**

votes

**1**answer

400 views

### Does Borel's proof for existence of normal numbers make an essential use of axiom of choice?

A normal number is a real number whose infinite sequence of digits in every base $b$ is distributed uniformly in the sense that each of the $b$ digit values has the same natural density $\frac{1}{b}$, ...

**2**

votes

**0**answers

118 views

### Defining Global Choice in terms of strong limit cardinals over $ZF$

In his answer to user33038's mathoverflow question "What axioms are stronger than the Axiom of choice?", Prof. Hamkins writes:
"What's more, the axiom of choice is equivalent over $ZF$ to the ...

**1**

vote

**0**answers

243 views

### Some questions regarding Shelah's revised Generalized Continuum Hypothesis [closed]

It is well known that $\mathsf{ZF}+\mathsf{GCH}\vdash\mathsf{AC}$ (which means that the Kunen inconsistency can be proven in $\mathsf{ZF}+\mathsf{GCH}$). Consider now Shelah's revised Generalized ...

**8**

votes

**0**answers

290 views

### A Banach-Tarski game

This is partially inspired by the question http://math.stackexchange.com/questions/1383397/cutting-a-banach-tarski-cake, which I find intriguing if unclearly written.
A paradoxical family of subsets ...

**4**

votes

**1**answer

138 views

### Does this axiom (a weak form of class valued choice) has a name?

At some point in my work (which has nothing to do with set theoretics foundation) I need to consider the following axiom:
For any set $X$, any class $V$ with a surjective map $f : V \...

**13**

votes

**1**answer

734 views

### Does “cardinal arithmetic is well-defined” imply axiom of choice?

Let me quickly explain what I mean with my question.
Let $(\kappa_i)_{i\in I}$ be a collection of cardinal numbers, indexed by elements of some set $I$. We can try to define $\sum\limits_{i\in I}\...

**1**

vote

**1**answer

97 views

### Freeness of the group of principal ideals of a number field

This is just a question wondering whether a concrete (enough) result that can be proved using the axiom of choice can be proved without it. The result being that the group of principal ideals $P_K$ of ...

**3**

votes

**1**answer

371 views

### A question about Cantor's Power Set theorem without the Axiom of Choice

Assume that we are working in ZF set theory without the Axiom of Choice. If S is an infinite set, let $S(f)$ denote the set of all finite subsets of $S$, let $S(I)$ denote the set of all infinite ...

**1**

vote

**1**answer

252 views

### The patterns of possibility for nontrivial automorphisms and nontrivial elementary embeddings of the universe

In their paper "The Role of the Foundation Axiom in the Kunen Inconsistency" (arXiv:1311.0814 [Math.LO]), Daghighi, Golshani, Hamkins, and Jerabek show that the patterns of possibility for the ...

**10**

votes

**3**answers

582 views

### Axiom of choice for sets of finite sets

The question I am going to ask is really to satisfy my curiosity, as I am not at all an expert of the subject and do not plan to really work on it. Hence, if you think the question is not suitable for ...

**7**

votes

**2**answers

776 views

### How do I apply the Boolean Prime Ideal Theorem?

I have become aware of an amazing phenomenon from a myriad of questions and answers here on MathOverflow: many of the results that I would typically prove using the Axiom of Choice can actually be ...

**6**

votes

**1**answer

296 views

### Logical strength of “choice functions exist for well-ordered families”?

A colleague of mine suggested the following weakening of the axiom of choice:
If $\mathscr{F} := \{F_\alpha\}$ is a well-ordered family of non-empty sets (i.e., there is a bijection between $\...

**6**

votes

**2**answers

288 views

### The role of the rigid relation principle ($RR$) in the Kunen inconsistency

Consider the rigid relation ($RR$) principle, i.e.
"every set admits a rigid binary relation", that is,"that for every set $A$ there is a binary relation $R$ on $A$ such that the structure $(A,R)$ is ...

**6**

votes

**1**answer

263 views

### Notions of infinity in $\mathsf{ZF}$ without choice

Consider the following statements about a given set $X$ in in $\mathsf{ZF}$:
(1) There is $x_0\in X$ such that there is a surjective map $\varphi: X\setminus\{x_0\}\to X$.
(2) There is an injective ...

**7**

votes

**2**answers

746 views

### Independence of the countable axiom of choice

How does one proove that the Countable axiom of choice is not provable in ZF?Is there any brief proof?Does the Independence of the countable axiom of choice implies the independence of the axiom of ...

**4**

votes

**0**answers

251 views

### Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$

Is it consistent with $\mathsf{ZF}$ (without $\mathsf{AC}$) that there is an infinite set $X$ and a subset $S\subseteq\mathcal P(X)$ of the same cardinality as $\mathcal P(X)$ with the property that ...

**0**

votes

**1**answer

167 views

### Noetherianess of a finite module over a noethrian ring without Axiom of Choice

All rings are assumed to be commutative with 1.
We say a module over a ring is strictly noetherian if every non-empty set of submodules has a maximal member. We say a ring is strictly noetherian if it ...

**2**

votes

**1**answer

110 views

### Matching power series to infinity

As pointed out by Makoto, on this question about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another ...

**6**

votes

**1**answer

266 views

### Are two forms of the Dual Schroeder-Bernstein property equivalent?

We know the Shroeder-Bernstein (SB) theorem can be proved in ZF, while the Dual Schroeder-Bernstein (DSB) can be proved in ZF+AC but not in ZF. Define as ISB the property that whenever there are both ...

**8**

votes

**1**answer

306 views

### Axiom of choice and vector spaces over a given field

It is my impression that the following question is open:
Does the existence of a basis for every vector space over the field K = the reals having a basis imply the axiom of choice?
I saw an answer ...

**2**

votes

**1**answer

382 views

### A question regarding the Hahn-Banach theorem

Wikipedia states that, in $ZF$, the Axiom of Choice ($AC$) implies the Hahn-Banach theorem, but that the Hahn-Banach theorem does not imply $AC$. It also states that in $ZF$, the Hahn-Banach theorem ...