An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty ...

learn more… | top users | synonyms

3
votes
1answer
103 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
1answer
139 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
1answer
171 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. ...
6
votes
1answer
308 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,...
1
vote
1answer
150 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
1answer
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 ...
28
votes
15answers
2k views

Objects which can't be defined without making choices but which end up independent of the choice

It happens a lot of times that when one defines a new object (ring, module, space, group, algebra, morphism, whatever) out of given data one first chooses some additional structure. And sometimes (...
4
votes
1answer
294 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, ...
14
votes
0answers
384 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 ...
22
votes
0answers
459 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
0answers
220 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
2answers
294 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 (...
59
votes
16answers
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 ...
7
votes
3answers
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 ...
2
votes
1answer
189 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 \...
7
votes
3answers
898 views

About the axiom of choice, the fundamental theorem of algebra, and real numbers

About fundamental theorem of algebra, there is a large collection different demonstrations. I ask: is there some proof that avoids AC (choice axiom)? In a general topos (with natural number object) ...
6
votes
1answer
214 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
1answer
283 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 ...
10
votes
1answer
406 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. ...
12
votes
0answers
281 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 ...
13
votes
1answer
479 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
2answers
365 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
0answers
94 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 ...
84
votes
2answers
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 ...
6
votes
1answer
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 ...
6
votes
0answers
222 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 ...
3
votes
0answers
307 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 ...
11
votes
1answer
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 ...
5
votes
1answer
600 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: ...
8
votes
1answer
367 views

Forcing, cuts, and Dedekind-finite cardinalities

Tl;dr version: there are two natural classes of cuts in the nonstandard model of arithmetic consisting of the Dedekind-finite sets (if, in fact, they constitute such a model); both these classes are ...
42
votes
4answers
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 ...
1
vote
1answer
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 ...
4
votes
0answers
174 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
1answer
396 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 $\frac{1}{b}$, ...
21
votes
2answers
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: ...
2
votes
0answers
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 ...
20
votes
0answers
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 ...
8
votes
0answers
286 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 ...
1
vote
0answers
242 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 ...
4
votes
1answer
136 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
1answer
726 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
1answer
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
5answers
944 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 ...
12
votes
1answer
546 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 ...
3
votes
1answer
364 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 ...
10
votes
3answers
578 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
2answers
759 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 ...
125
votes
15answers
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 ...
6
votes
1answer
293 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
2answers
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 ...