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

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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 ...

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### 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 ...

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### 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 ...

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### 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 $P$ ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### Does ZFC prove the universe is linearly orderable?

It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example ...

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### Does ZF bound countable unions of countable sets?

ZF proves that whenever a countable union of countable sets can be well ordered then its cardinality is at most $\aleph_1$. But what if it cannot be well ordered? The Feferman-Levy model shows the ...

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### Inequivalent complete norms and the axiom of choice

Hi,
I've been wondering about the following :
Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space?
All the examples of inequivalent complete norms ...

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### Kaplansky's theorem and Axiom of choice

Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow:
Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...

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### In ZF, when is a disjoint union of metrizable spaces metrizable?

It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of
metric spaces is metrizable, simply by rescaling or chopping off
the individual metrics to have diameter at most one, and ...

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### A combinatorial property implied by the Axiom of Choice

Let us say that a family $R$ of sets has the Finite Subcovering Property --- FSP --- if any subfamily of $R$ which covers the union $\cup B: B \in R$ has itself a finite subfamily which also covers. ...

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### Existence of algebraic closure and Axiom of choice [duplicate]

Possible Duplicates:
Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
algebraic closure of commuting pairs of matrices
we need ...

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### Without choice, can every homomorphism from a profinite group to a finite group be continuous?

In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the ...

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### On surjections, idempotence and axiom of choice

The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF?
(I couldn't find it in the Consequences site so far.)
If $A$ is an infinite set such that $A$ ...

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### What is a Choice Principle, really?

This question is quite soft, and I apologize in advance if it borderline off-topic.
When working in theories between ZF and ZFC the term "choice principle" is heard quite often. For example:
$\quad$ ...

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### Strictly order preserving maps into the integers

If $P$ and $P'$ are partial orders, a strictly order preserving map from $P$ to $P'$ is an $f:P\to P'$ satisfying that $x\lt y$ implies $f(x)\lt f(y)$ for all $x,y\in P$.
An interval in $P$ is a set ...

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### $\Theta$ and the Hartogs of $2^\mathbb R$

Let $a,b$ be sets, we write $a\leq^\ast b$ if either $a=\varnothing$ or there exists a surjection $f\colon b\to a$. With the axiom of choice this is a linear ordering equivalent to the usual ordering ...

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### 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 ...

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### Distinguishing two local versions of the axiom of choice

Two equivalent formulations of the axiom of choice are:
Every family $(X_i)_{i \in I}$ of pairwise disjoint nonempty subsets of a set $X$ has a choice function.
Every family $(X_i)_{i \in I}$ of ...

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### Minimal blocks for a family of finite sets

In this question I asked for a reference for the following lemma:
Lemma X: For every family $\mathcal G$ of nonempty finite sets there
is a minimal "blocking set" $B$. By a "blocking set" $B$ I ...

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### Minimal selector for a family of finite sets

A colleague is refereeing a paper in which the following
lemma appears implicitly:
For any family $\mathcal G$ of nonempty sets let us call
a set $B$ a "selector" if $B$ meets all $F\in\mathcal ...

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### Bounded operators and axiom of choice

In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of ...

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### Minimal prime ideals and Axiom of Choice(revised version)

From the page:
Existence of prime ideals and Axiom of Choice.,
I have found that The existence of prime ideals in commutative rings is equivalent to the Boolean Prime Ideal theorem. But $BPI$ is ...

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### Existence of prime ideals and Axiom of Choice.

One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple ...

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### Hartogs number and the three power sets

One of the most important constructions in ZF+$\lnot$AC is Hartogs number, defined as:
$$\aleph(X)=\min\lbrace\alpha:|\alpha|\nleq|X|\rbrace$$
We can prove that this ordinal always exists in the ...

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### Axiom of Choice and Vitali's theorem [duplicate]

Possible Duplicate:
Axiom of choice and non measurable set
I am told somebody has shown the equivalence of the Axiom of Choice with
existence of non-measurable Lebesgue sets on the real line, ...

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### Logical relationships between weakenings of AC

What are the known logical implications between weak choice principles like $DC_\kappa$", the ...

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### Counterexemple to Urysohn's lemma in a topos without denombrable choice ?

Hello !
The Urysohn's Lemma assert that in every topological spaces which is normal two closed subset may be separated by a real valued function. It's proof use axiom of countable choice (but not the ...

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### Supercompact and Reinhardt cardinals without choice

A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:
Definition. A cardinal $\kappa$ is supercompact if for all ordinals ...

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### Is choice needed to establish the existence of idempotent ultrafilters?

It is well known that the Stone–Čech compactification $\beta \mathbb N^+$ of the positive natural numbers has the structure of a compact left semitopological semigroup and hence, by Ellis's lemma, has ...

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### Concerning the various proofs from the axiom of choice that R^3 admits of surprising geometrical decompositions into circles, skew lines and so on: can we prove in any instance that there are no Borel such decompositions? Or that AC is required?

This question follows up on a comment I made on Joseph O'Rourke's
recent question, one of several questions here on mathoverflow
concerning surprising geometric partitions of space using the axiom
of ...

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### Sizes of bases of vector spaces without the axiom of choice

Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two ...

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### Countably generated $\sigma$-algebras of ${\mathcal P}({\mathbb R})$ and choice

It is consistent with ${\sf ZF}$ that the reals are the countable union of countable sets. Since any countable set is Borel, it follows that in any such pathological universe, let's call it $W$, every ...

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### On Successive Regular Cardinals With No Ladders

Definition: Let $\kappa$ be an $\aleph$ cardinal, we say that $\langle f_\alpha\colon\alpha\to\kappa\mid\alpha<\kappa^+\rangle$ is a ladder if every $f_\alpha$ is injective.
Equivalently this is ...

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### What sort of large cardinal can $\aleph_1$ be without the axiom of choice?

Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal.
...

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### 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 ...