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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
164 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
879 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
187 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
253 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
334 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. ...
10
votes
0answers
244 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 ...
1
vote
0answers
148 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, ...
12
votes
1answer
435 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
333 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
77 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 ...
81
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
171 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
207 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
0answers
242 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 ...
3
votes
0answers
289 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 ...
6
votes
1answer
567 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
347 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 ...
39
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
232 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
164 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 ...
2
votes
1answer
376 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 $\frac{1}{b}$, ...
20
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
117 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
267 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
221 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
130 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
658 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 ...
1
vote
1answer
95 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
898 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
535 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
326 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
557 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
704 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 ...
121
votes
15answers
23k 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
285 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
286 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
1answer
255 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 ...
10
votes
4answers
724 views

On surjections, idempotence and axiom of choice

The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF? (I couldn't find it in the Consequences site so far.) If $A$ is an infinite set such that $A$ ...
7
votes
2answers
661 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
0answers
244 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 ...
16
votes
1answer
755 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 ...
0
votes
1answer
161 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 ...
9
votes
2answers
408 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 ...
2
votes
1answer
107 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 ...
8
votes
1answer
673 views

Can we prove that the ring of formal power series over a noetherian ring is noetherian without axiom of choice?

Let $A$ be a commutative ring with an identity. Suppose that every non-empty set of ideals of $A$ has a maximal element. Let $A[[x]]$ be the formal power series ring over $A$. Can we prove that every ...
12
votes
5answers
878 views

Does k(X) have a k-basis for every set X, without AC?

This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?. For any field $k$, the field $k(x)$ of rational functions in one variable has an ...
6
votes
1answer
258 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 ...