The tag has no wiki summary.

learn more… | top users | synonyms

8
votes
0answers
443 views

Seemingly elementary geometric problem in R^3 which requires the axiom of choice

While playing with what I called "quantum matching", the following problem arose: construct a map $F$ from the unit sphere $S_2$ in $R^3$ to itself such that $F(X)$ is orthogonal to $X$ plus has one ...
7
votes
1answer
387 views

Non continuous Linear form on $E=C([0,1],\mathbb{R})$ without AC

Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm. Is it possible to show a non-continuous linear form on $E$ exists ...
2
votes
0answers
60 views

Metric space has a basis countably locally finite

it is know that all metric space has a basis countably locally finite and this result is proved by using axiom of choice. Then, the natural question is: is possible to prove this result without using ...
2
votes
1answer
215 views

What is the order type of $L$ with Godel's well ordering?

In some sense $Ord$ is a "proper class" ordinal. Unfortunately the notion of a proper class ordinal is not a straight forward generalization of the notion of "set" ordinals because the proper classes ...
5
votes
0answers
177 views

Proving equivalence of a tree-based version of Countable Choice for families of finite sets

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
17
votes
1answer
700 views

Linear Algebra without Choice

We consider the field of "usual" linear algebra. Q. Which aspects of it can be carried out without the Axiom of Choice? Q. Do interesting "exotic" phenomena appear in presence of (some instance of) ...
12
votes
0answers
354 views

How much choice is required to prove concretizability theorems in category theory?

A concretization of a category is a faithful functor to the category of sets. A category is concretizable if there exists such a functor. An evident necessary condition for concretizability is ...
11
votes
1answer
229 views

Pullback-stability of internally projective objects

An object $X$ of a category $C$ is said to be projective if the hom-functor $C(X,-)$ preserves epimorphisms (or, in general, some restricted class of epimorphisms such as the regular or effective ...
14
votes
1answer
374 views

Does ZF imply a weak version of Hahn-Banach?

I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple ...
7
votes
2answers
248 views

How to make countably closed forcing “nice” without choice

When working over a model $V$ of $ZFC$, countably closed forcings are extremely nice: If $\mathbb{P}$ is countably closed, then $V[G]$ has no new $\omega$-sequences of elements of $V$. In ...
1
vote
1answer
101 views

Some definitions without full choice

Assume DC($\aleph_1$). Can we define the following: Basis for a vector space $V$ over a field $K$ such that $\operatorname{card}(K) \leq \aleph_1$ and we happen to find a generating set of $V$ of ...
10
votes
2answers
267 views

Singular successors without large cardinals

Assuming the axiom of choice we have that successor cardinals are regular. However as one of the first examples of uses of forcing show, it is consistent relative to $\sf ZF$ that $\omega_1$ is ...
2
votes
1answer
131 views

Axiom of dependent choice (up to $\omega_1$) and group rank

Assuming the axiom DC($\omega_1$), is there a definition of the rank of a group ? Another related question: assuming DC($\omega_1$), if we have two groups $A$ and $B$ of the same infinite rank, is ...
5
votes
2answers
163 views

A few standard results (on metrizability and relative separation strength) without Choice?

I've been going back over some results from Munkres's Topology, and I'm curious about some things. (I originally posted this on M.SE, but I think it is probably a better fit here.) I know that Choice ...
9
votes
0answers
177 views

The global dimension of fields

In the absence of the Axiom of Choice, it is not necessarily true that all vector spaces over a field have bases. What are the possible global dimensions of fields in a model of ZF in which AoC ...
8
votes
1answer
481 views

A question about the Axiom of Choice

Let AC denote the Axiom of Choice. Let PP denote the so-called "Partition Principle" which states that "If S is a non-empty set and T is a non-empty set of pairwise disjoint subsets of S, then S can ...
5
votes
1answer
212 views

Proper-class sized “ring” with no maximal ideals

Suppose I have a collection of "elements" together with operations that satisfy the axioms for a commutative ring with identity --- except that these elements form not a set, but a proper class. Must ...
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 ...
4
votes
1answer
191 views

Well-Ordering theorem of cardinal$\kappa$

I've heard from others about the WO($\kappa$) as a counterpart of AC($\kappa$), but I cannot find a suitable way to express it in ZF since "every set of cardnality $\kappa$ can be well-ordered" is ...
9
votes
1answer
301 views

Cardinals without choice: interpolation (reference wanted)

Is there a published reference for this ZF theorem? Let $m,n\in\mathbb{N}$. If $a_1,\dots,a_m$ and $b_1,\dots,b_n$ are cardinals such that $a_i\le b_j$ for all $i$ and $j$, then there is a cardinal ...
8
votes
1answer
421 views

Weakest choice principle required for Robertson-Seymour Graph Minor Theorem?

The main Robertson-Seymour Theorem states that finite graphs form a well-quasi-ordering under the graph minor relation. In other words, in every infinite set of finite graphs, there exist two graphs ...
21
votes
2answers
1k views

Hahn's Embedding Theorem and the oldest open question in set theory

Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his paper Über die nichtarchimedischen Grössensysteme (Sitzungsberichte der ...
2
votes
0answers
308 views

Is the axiom of choice really related to choice? [closed]

I am not an an expert in set theory, so this question could be trivial. I am sorry in that case. Let $I$ be a set and $\{ X_i \}_{i \in I}$ be a collection of sets such that $X_i \neq \emptyset$ for ...
3
votes
1answer
348 views

How much of ZFC do I need to construct this cofinal, order-preserving class function?

EDIT: I'm bumping this, because while Joel ruled out some naive options, my question in bold below is not yet answered. Suppose I have a directed partially ordered set $(\Gamma,\leq)$ with a bottom ...
13
votes
1answer
852 views

Non-constructive existence proofs without AC?

Hi everyone, This is a question I have been asking from long, but none of my colleagues could ever answer me: It is a well-known fact that the axiom of choice (AC) allows one to prove the existence ...
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 ...
4
votes
2answers
680 views

Are all models of ZF + DC + “All set of reals are lebesgue measurable” also models of CH? [duplicate]

Possible Duplicate: Lebesgue Measurability and Weak CH I have studied a little set theory and I found that Solovay constructed a model of ZF+DC+"All set of reals are Lebesgue measurable" and ...
3
votes
3answers
462 views

Set theory question

It is known from a result of Sierpinski that the generalized continuum hypothesis (GCH) implies the axiom of choice (AC). It is also known from the celebrated results of Cohen that AC is independent ...
11
votes
3answers
1k views

Cantor's diagonal argument and ZF

Cantor's diagonalization construction, on a certain view, furnishes functions $$d_X:{\rm Injections}(X,P(X))\rightarrow P(X)$$ that satisfy $\forall X\forall i\ \ d_X(i)\not\in i(X)$ In ZF, can ...
8
votes
3answers
795 views

Axiom of Choice and continuous functions

Do you know if the following statement is an equivalent form of the axiom of choice or not? If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is ...
2
votes
0answers
261 views

on the Axiom of Choice and the Spectrum of Rings

consider the following theorem, when $R$ is a commutative ring with a non-zero identity: A ring $R$ is zero-dimensional if and only if $\mbox{Spec(R)}$ is Hausdorff. The proof uses the Axiom of ...
2
votes
2answers
567 views

Mathematics with the negation of AC

Clearly Very important results in Math require the Axiom of choice, for example "any vector space has a base". But in the absence of AC (i.e., only in ZF) it is possible that a vector space has no ...
7
votes
2answers
654 views

Category and the axiom of choice

What are (if any) equivalent forms of AC (The Axiom of Choice) in Category Theory ?
2
votes
1answer
346 views

Notation arb(x)

Suppose we have extended $ZF$ by adding to $ZF$ an unary function symbol $arb$ (an arbitrary element of a set) and a corresponding axiom "For every non-empty set $S$, $arb(S)$ is in $S$". Will be the ...
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?
0
votes
2answers
242 views

Equivalent Forms of AC

There are many algebraic equivalences of AC in the literature. A famous one is "every ring with identity has a maximal ideal". Where can I find this equivalences, specially those in rings theory !? ...
2
votes
2answers
282 views

what axioms are between AC and Countable choice !

Hi All. Need some information. We all know axiom of choice (AC) and countable choice. Which axioms are between these two. I mean weaker that Axiom of choice but stronger than countable choice ?
5
votes
1answer
308 views

Existence of model of ZF without AC, but with many choice function

Question 1: Does there exist models of the Zermelo-Fraenkel set theory without the axiom of choice, but such that every indexed family of non-void sets whose index set has a well-orderable cardinal ...
4
votes
5answers
1k views

What axioms are stronger than the Axiom of choice?

What other axioms in set theory are stronger than AC ? I mean what are those axioms that will imply AC ?
2
votes
3answers
267 views

Construct a fixed-point set operator

How to find an uncountable set $S$, and construct an function $f : 2^S \longrightarrow S$ such that for any $T \subseteq S$, $f \left( T \right) \in T$? for example, let $S =\mathbb{R}$, how can I ...
5
votes
1answer
238 views

Is this height-transcendence-degree inequality true without AC ?

Let $R$ be a $k$-algebra ($k$ a field) and a domain of finite Krull dimension. In $\quad$ Krull dimension <= transcendence degree? it is shown that $$\text{Krull-dim}(R) \le \text{trans.deg}_k ...
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 ...
5
votes
1answer
300 views

Name for this generalized pigeonhole principle?

For a set $X$, let $|X|$ denote its cardinality. A block of a partition is a non-empty element of the partition. Let $P$ and $Q$ be two partitions of a set $X$. If $|P| < |Q|$ then $P$ ...
5
votes
8answers
720 views

Result that follows from ZFC and not ZF but are strictly weaker than choice

A number of results that people use that require the axiom of choice (i.e. do not follow from ZF alone) are known to actually imply the axiom of choice. Therefore, one might naturally wonder whether ...
4
votes
4answers
665 views

A question about the Axiom of Choice and straight lines in the Euclidean plane.

Let E be the Euclidean plane. Does there exist a collection C of subsets of E whose union is E and which are all straight lines such that (1) No two distinct straight lines belonging to C are parallel ...
15
votes
1answer
496 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 ...
5
votes
1answer
225 views

Complete anti-chain lattices and the axiom of choice

Hello, everyone. I'm trying to find out about lattices of anti-chains, and was wondering whether you could help me with getting to grips with a Comp. Sci. paper I'm struggling with. I've been reading ...
12
votes
1answer
739 views

Differential equations and axiom of choice

In the most general context, the Picard-Lindelöf theorem (aka Cauchy-Lipschitz in French) asserts the existence of a maximal solution for $\dot{x}(t) = f(t,x(t))$, i.e. of a solution $x(t)$ defined on ...
8
votes
1answer
363 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 ...
5
votes
0answers
355 views

Existence of Non-Borel sets in models of “All sets measurable”

We know that the consistency of ZFC+"Exists an inaccessible cardinal" implies the consistency of ZF+DC+"All sets are Lebesgue measurable"; and DC proves the existence of non-Borel sets. J. Truss ...