The axiom-of-choice tag has no wiki summary.

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### Probabilities in a riddle involving axiom of choice

The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question):
The Riddle:
We assume ...

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### 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: ...

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**1**answer

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### Subsets of Real Numbers (Edited & Revised Version)

Question 1: Is it consistent with $\text{ZF}$ that only countable subsets of $\mathbb{R}$ are well-orderable?
Question 2: Is it consistent that for some $\lambda$, $\aleph_0 < \lambda < ...

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### Minimal Generalized Continuum Hypothesis & Axiom of Choice

It is well known that working in the frame of $\text{ZF}$, the Generalized Continuum Hypothesis ($\text{GCH}$) implies the Axiom of Choice ($\text{AC}$), i.e. $\text{ZF}+\text{GCH}\vdash \text{AC}$.
...

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### Discontinuous representations of GL(n,C) in ZF

Discontinuous linear representations of $GL(n,\mathbb{C})$ can be obtained from the so-called "wild" (field) automorphisms of $\mathbb{C}$; but these wild automorphisms in turn require some choice to ...

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### Full conditional probabilities and versions of AC?

A probability is a finitely additive measure on a boolean algebra with total measure $1$.
A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean ...

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401 views

### Strength of some claims about finitely additive measures on infinite sets?

Assume ZF. Consider the claim:
(1) For any infinite set $\Omega$, there is a finitely additive probability measure $\mu:2^\Omega\to[0,1]$ with $\mu(A) = 0$ whenever $|A|<|\Omega|$.
Then (1) is ...

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246 views

### Possible Choices for Cofinality of $\aleph_n$ without Choice

$\text{ZFC}$ proves that each $\aleph_{n}$ for $n\in \omega$ is a regular cardinal. But it seems without the Axiom of Choice there are many consistent possible choices for cofinality of such ...

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437 views

### Is sigma-additivity of Lebesgue measure deducible from ZF?

Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)?
Note 1. ...

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### 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 ...

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### 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 ...

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393 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 ...

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61 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 ...

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222 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 ...

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### 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 ...

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742 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) ...

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359 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 ...

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### 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 ...

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402 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### Category and the axiom of choice

What are (if any) equivalent forms of AC (The Axiom of Choice) in Category Theory ?

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### 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 ...

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### 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?

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### 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 !? ...

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### 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 ?

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### 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 ...

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### 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 ?

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### 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 ...