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1
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1answer
121 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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0answers
102 views

Existence and local compactness of the p-adic number field without Axiom of Choice [closed]

I think we can prove the existence and local compactness of the p-adic number field without using Axiom of Choice. Am I right?
9
votes
2answers
390 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 ...
34
votes
4answers
8k 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 ...
2
votes
1answer
90 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 ...
7
votes
1answer
493 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 ...
11
votes
5answers
769 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
209 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
272 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
334 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 ...
8
votes
1answer
247 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
223 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
236 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 ...
8
votes
5answers
2k views

Subset of the plane that intersects every line exactly twice

In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to ...
4
votes
1answer
360 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 ...
6
votes
2answers
243 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 ...
15
votes
5answers
1k views

Compactness of the Hilbert cube without the Axiom of Choice

I am just curious: is there a published proof of the compactness of the Hilbert cube that does not use the Axiom of Choice, or is it well known?
7
votes
1answer
217 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
318 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
170 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
1k 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 ...
15
votes
1answer
514 views

Categorifications of Zorn's lemma

I'm wondering about categorifications of Zorn's lemma along the following lines. Lemma: if $\mathbf{C}$ is a small category in which every directed diagram of monomorphisms has a cocone of ...
46
votes
15answers
12k 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 ...
3
votes
1answer
164 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
198 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 ...
21
votes
2answers
1k views

Axiom of Choice: Ultrafilter vs. Vitali set

It is well known that from a free (non-principal) ultrafilter on $\omega$ one can define a non-measurable set of reals. The older example of a non-measurable set is the Vitali set, a set of ...
9
votes
1answer
519 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 ...
25
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 ...
7
votes
1answer
285 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 ...
16
votes
4answers
1k views

What are the known implications of “There exists a Reinhardt cardinal” in the theory “ZF + j”?

This is, alas, in large part a series of questions on unpublished work of Hugh Woodin; it's also quite frivolous if Reinhardt cardinals turn out inconsistent. Definitions: Call $\kappa$ an ...
6
votes
2answers
399 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
267 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 ...
3
votes
1answer
163 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
213 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 ...
6
votes
1answer
223 views

Solovay's Theorem on Partitions of Stationary Sets and Weak Choice Principles

There is a weak choice principle called $DC_\lambda$ which holds in $L(V_{\lambda+1})$ under the assumption of a non-trivial elementary embedding $$j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ and it is ...
14
votes
1answer
948 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, ...
5
votes
1answer
365 views

Comparability implies well-orderability?

I am trying to prove a small proposition that got me completely stumped, and I cannot find a single counterexample. (ZF) Suppose that $E$ is such that for every $A\subseteq\mathcal P(E)$ either ...
37
votes
1answer
1k views

When does $A^A=2^A$ without the axiom of choice?

Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$ However without the axiom of choice this doesn't ...
6
votes
0answers
200 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
255 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
223 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
610 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 ...
7
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0answers
143 views

Strength of claims about extensions of partial preorders and orders to linear ones

Consider these two axioms: Every partial order extends to a linear order. Every partial preorder (reflexive and transitive relation) extends to a linear preorder while preserving strict orderings: ...
5
votes
3answers
228 views

Well-ordering with a topological property

Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it. An ...
2
votes
0answers
379 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
225 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
399 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
759 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 ...
14
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2answers
516 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 ...