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6
votes
0answers
108 views

What is the significance of having Prime Ideal Theorem in models for failure of Axiom of Choice? [migrated]

Prime Ideal Theorem says: PIT: Every ideal on a Boolean algebra can be extended to a prime ideal. It follows from Axiom of Choice but is weaker than it. In many cases I saw that people check ...
3
votes
1answer
243 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
0answers
174 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
3answers
498 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 ...
6
votes
2answers
606 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
261 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 ...
5
votes
2answers
265 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 ...
8
votes
1answer
241 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
2answers
529 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 ...
6
votes
0answers
233 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 ...
1
vote
1answer
144 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
1answer
97 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
1answer
225 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
1answer
284 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
1answer
351 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 ...
12
votes
5answers
807 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
votes
1answer
263 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
votes
0answers
241 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
votes
0answers
258 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
votes
2answers
268 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
votes
1answer
237 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
votes
1answer
362 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
votes
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 ...
7
votes
0answers
183 views

Countable choice in $L(\mathbb{R}^*_G)$

Let $\lambda$ be a singular strong limit cardinal and let $G \subset \text{Col}(\omega,\mathord{<}\lambda)$ be a $V$-generic filter. Let $\mathbb{R}^*_G = \bigcup_{\alpha < \lambda} ...
13
votes
4answers
2k views

Is it possible to formulate the axiom of choice as the existence of a survival strategy?

Consider the following situation: There is an infinite set $G$ of giraffes. A lion comes and announces a set $C$ of all possible colours and an infinite cardinal $\kappa$. The hungry lion ...
3
votes
1answer
176 views

Does existence of $\omega_1$ subset of reals imply $\omega_1$ choice for subsets of reals?

Suppose there exists a subset of $\Bbb R$ which has cardinality $\omega_1$. Is it then necessarilly true that for every collection of $\omega_1$ subsets of $\Bbb R$ there exists a choice function? I ...
9
votes
1answer
541 views

Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?

Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial(scaler multiplication is not identicaly zero) $\mathbb{C}$-module. Now my questions are? 0.Is it consistent with $ZF$ that $\mathbb{R}$ is ...
6
votes
2answers
456 views

Exponentiation and Dedekind-finite cardinals

It is known that the sum and the product of two Dedekind-finite cardinals are also Dedekind-finite cardinals. What about cardinal exponentiation ? Question: Let A and B be two Dedekind-finite ...
5
votes
1answer
298 views

Relationship between fragments of the axiom of choice and the dependent choice principles

The dependent choice principle ${\rm DC}_\kappa$ states that if $S$ is a nonempty set and $R$ is a binary relation such that for every $s\in S^{\lt\kappa}$, there is $x\in S$ with $sRx$, then there ...
9
votes
1answer
201 views

Without AC, which implications between the different definitions of amenability still hold?

More precisely, I would like to know which implications between the following definitions of amenability of a discrete countable (or even finitely generated) group can be proved to hold with only ZF ...
3
votes
1answer
168 views

Is it compatible with ZF to assume that every amenable discrete group is finite?

The question is in the title, amenability being understood as the existence of a left-invariant finitely additive probability measure on the group of interest. The case of countable groups is treated ...
5
votes
2answers
219 views

Forcing $\neg AC$

Sorry if this sounds like a silly reference request, but I wasn't able to track down any. I'm looking for proof, via forcing, that axiom of choice can fail in a model of $ZF$. All of papers I found ...
7
votes
1answer
299 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 ...
14
votes
1answer
1k views

How much of GCH do we need to guarantee well-ordering of continuum?

It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, ...
6
votes
0answers
210 views

Does Sageev's result need an inaccessible?

In 1981, building on work by Ellentuck in 1974, Sageev showed ("A model of ZF + there exists an inaccessible, in which the Dedekind cardinals constitute a natural non-standard model of arithmetic," ...
4
votes
1answer
268 views

What axioms (other than choice) have a taming effect on the ordering of cardinalities?

Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, ...
4
votes
1answer
225 views

Choice principle strong enough to defy $V=L$

When using axiom of choice in proofs, people often say that this is non-constructive because AC gives us only proofs of existence, without giving explicit example. However, because in $L$ AC holds, we ...
21
votes
1answer
622 views

Is there a model of ZF set theory with a set that does not inject into the cardinals?

Question. Is there is a model of ZF set theory with a set $X$ that does not inject into the cardinals? I use the term "cardinal" here in the ZF sense, so they are not necessarily well-orderable. To ...
2
votes
0answers
395 views

Can we prove an open affine subscheme of a noetherian scheme is noetherian without Axiom of Choice?

I'm interested in proving basic results of algebraic geometry without Axiom of Choice. As for why I think this is interesting, please see Pete L. Clark's answer to this question. To state my problem, ...
6
votes
1answer
229 views

Sets of cardinalities of bases without choice

For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...
4
votes
2answers
416 views

Properties of vector spaces without AC

With AC, it is easy to see that any vector space is injective, and free, therefore alse flat and projective. Without AC, vector spaces can be not free. Are they must be projective modules? Flat ...
3
votes
1answer
764 views

Can we construct cohomolgy theory on noetherian separated schemes without Axiom of Choice?

The usual cohomology theory on schemes uses injective or flasque resolutions of quasi-coherent sheaves. Hence it uses Axiom of Choice. However, if the base scheme is a noetherian separated scheme, the ...
16
votes
0answers
327 views

Are the reals really a fraction field?

In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$. A crucial point in the ...
14
votes
2answers
532 views

Pathological behavior of Borel sets?

Usually in set theory, Borel sets are much more nicely behaved than arbitrary sets of reals. One reason for this is Borel determinacy, which immediately yields measurability, Baireness, and the ...
5
votes
4answers
331 views

Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?

This question comes after the comments in the recent related question Sigma-complete Lindenbaum algebras?, but in its current form is sufficiently different in my opinion, and so I decided to follow ...
1
vote
3answers
316 views

Why doesn't choice imply global choice (in NBG)?

I thought ZFC proved the existence of an inductive well-ordering that is itself a set for any stage of V. NBG with only the regular AC should then prove/assert the existence of a class R of ordered ...
17
votes
1answer
278 views

Linear maps between arbitrarily chosen vectors of vector spaces $V$ and $W$

I recently came across this question: Is the axiom of choice needed to prove the following statement: Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. ...
5
votes
1answer
156 views

Function Approximation in c.c.c Forcings without AC in Ground Model

Consider the following basic theorem. Theorem. If $M$ is a c.t.m of ZFC and $\mathbb{P}$ a c.c.c forcing notion in $M$ and $G$ a $\mathbb{P}$ - generic filter on $M$ then for all $A,B$ in $M$ and for ...
4
votes
1answer
371 views

Re-interpreting vector spaces in a choice-less model of ZF as modules over a regular ring in ZFC

I am searching a module $M$ over a (von Neumann) regular ring $A$ ($\forall a\in A$, $\exists x\in A$: $axa=a$) with two properties: (1) every finitely generated submodule of $M$ is projective ...
5
votes
1answer
167 views

Intermediate submodels which do not satisfy AC

The following is known: Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of ...