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 ...

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126 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
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344 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
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1answer
460 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 ...
22
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0answers
454 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 ...
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213 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
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2answers
291 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
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1answer
185 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 ...
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1answer
213 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
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1answer
281 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
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273 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
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1answer
390 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. ...
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1answer
276 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
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1answer
473 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
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2answers
359 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 ...
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0answers
92 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
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1answer
189 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
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0answers
218 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
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0answers
252 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
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0answers
303 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
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1answer
593 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: ...
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173 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
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1answer
394 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
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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 ...
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239 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 ...
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275 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
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1answer
134 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
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1answer
723 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 ...
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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
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1answer
357 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 ...
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1answer
250 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 ...
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3answers
577 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
750 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
1answer
291 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
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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 ...
6
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1answer
261 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 ...
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2answers
725 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
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0answers
249 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
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1answer
165 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
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1answer
109 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
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1answer
262 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
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1answer
303 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
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1answer
378 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 ...
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5answers
899 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 ...
8
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1answer
280 views

What is known about global well ordering of classes in Gödel-Bernays?

I would like to have something like a linear order on classes, such that every instantiated predicate of classes has a minimal instance in that order. For my purposes, it is fine to assume V=L for ...
6
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0answers
263 views

A new cardinality living in every forcing extension?

This question is motivated by the papers http://arxiv.org/abs/1405.7456 and http://arxiv.org/abs/1410.1224. Say that a set $X$ is "generically presentable" over $V\models ZF$ if there is some ...
5
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0answers
288 views

Cardinal characteristics without choice

(I'm taking my definition of a cardinal characteristic from Blass' excellent article http://www.math.lsa.umich.edu/~ablass/need.pdf, which cites Vojtas/Fremlin/Miller; theirs is more general, but I'm ...
6
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2answers
306 views

When does Skolemization require the axiom of choice?

Skolemization is often used for eliminating existential quantifiers, which is often useful for proving theorems, especially in automated resolution theorem proving. Skolemization in first order ...
7
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1answer
265 views

Are there known ways to posit definable global choice in ZF without positing V=L?

I need a global choice function defined by a formula in (a fragment of) ZF. There is no harm in assuming V=L for my purposes. But I wonder if there are any familiar alternative ways to get this? ...
11
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1answer
470 views

What are the current views on consistency of Reinhardt cardinals without AC?

It's well known that Reinhardt cardinals are inconsistent, provided that we have access to axiom of choice, but, as far as I know, we are clueless about this when we don't assume choice. For me, the ...
11
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3answers
1k views

Difference between ZFC and ZF+GCH

I hear that the axiom of choice (AC) derives from The generalized continuum hypothesis(GCH). And also hear that both AC and GCH are independent of Zermelo–Fraenkel set theory(ZF). So, I'm just ...