The axiom-of-choice tag has no usage guidance.

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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 ...

**4**

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**1**answer

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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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262 views

### 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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**4**answers

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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This is, alas, in large part a series of questions on unpublished work of Hugh Woodin; it's also quite frivolous if Reinhardt cardinals turn out inconsistent.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 ...

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**2**answers

504 views

### Axiom of choice and convergence

Hi fellows,
I was wondering. Is the axiom of choice used to show that $\mathbb{R}$ is complete? If yes, is there a way to construct monotonic bounded sequences that do not converge?
Thanks in ...

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**2**answers

439 views

### What sort of structure can amorphous sets support?

Assuming the Axiom of Choice, every cardinal is either finite (i.e., an element of $\omega$) or Dedekind-infinite (i.e., in bijection with a proper subset of itself). This dichotomy is not true in ZF, ...

**17**

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### The Continuum Hypothesis and Countable Unions

I recently edited an answer of mine on math.SE which discussed the implication of the two assertions:
$AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and
$CH$ which says that if $A\subseteq 2^{\omega}$ ...

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**3**answers

543 views

### Surjective Maps onto $\aleph$-numbers

We denote by $\frak p\le q$ the abbreviation that there is $f:\frak p\to q$ which is injective, and by $\frak p\le^\ast q$ we abbreviate that there is a surjection from $\frak q$ onto $\frak p$.
If ...

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**0**answers

271 views

### What are these sets in Freyd's model?

Recall Freyd's model of $ZF +\neg AC$ (as recounted in MacLane and Moerdijk's book Sheaves in Geometry and Logic): it arises as the Fourman interpretation of the topos of sheaves on a particular ...

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562 views

### Antichains of Cardinals in ZF Without Choice…

With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what ...

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338 views

### New Foundations and weaker forms of choice

New Foundations (introduced by Quine) proves that $AC$ is false. Out of curiosity, is $NF$ consistent with countable choice or dependent choice? What's the strongest consequence of choice still ...

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**1**answer

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### Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice?

The Generalized Continuum Hypothesis can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, The Axiom of Choice, Theorem 9.1 p.133).
In fact, a relatively weak ...

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### Unique Existence and the Axiom of Choice

The axiom of choice states that arbitrary products of nonempty sets are nonempty.
Clearly, we only need the axiom of choice to show the non-emptiness of the product if
there are infinitely many ...

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762 views

### Two questions about vector spaces absent AC.

My questions are motivated by
this question
which asks, in the absence of AC, whether a subspace of a vector space with a basis must have a basis.
Does every real vector space embed isomorphically ...

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**1**answer

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### If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?

Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if:
Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
If $\sum\alpha_i b_i = 0$, where ...

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801 views

### Does every group of order bigger than 2 have a non-trivial automorphism?

If $G$ is a non-abelian group, then it has a non-trivial inner automorphism (conjugation by any non-central element). If $G$ is abelian of exponent bigger than 2, then the inversion map is an ...

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488 views

### Example of a topos that violates countable choice

At this nLab page we have the line
In contrast, any topos that violates countable choice, of which there are plenty, must also violate internal COSHEP.
It doesn't give an example, and neither ...

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517 views

### Forcing the nonexistence of a certain set

I have a certain set-theoretic axiom (WISC) which follows from Choice (this is a nuking a fly BTW), but which I suspect is independent of ZF. To show this I need to show that a certain set does not ...

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**1**answer

584 views

### Is it consistent relative to ZF that $\frak c = \aleph_\omega$?

In ZFC we know that the continuum cannot have cofinality $\omega$.
However, in the Feferman-Levy model we have that $\frak c=\aleph_1$, and that $\operatorname{cf}(\omega_1)=\omega$. In fact in the ...

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**1**answer

236 views

### Is the ordering principle preserved in generic extensions?

The ordering principle says that every set can be linearly ordered.
In a previous question Why are some axioms preserved in generic extensions? Asaf Karagila asked which axioms are preserved in ...

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**2**answers

661 views

### Why are some axioms preserved in generic extensions?

It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$.
On the other hand, it is long known that ...

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### Forcing over models without the axiom of choice

In the vast majority of papers forcing is always developed over ZFC.
Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain ...

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**1**answer

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### Countable Unions And The Axiom Of Countable Choice

Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of ...

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**1**answer

404 views

### Symmetric extensions and class forcing

Suppose $V\models ZFC$ and $P\in V$ is a poset of forcing conditions.
It is a basic theorem in forcing that $V[G]\models ZFC$ for any generic extension by a $V$-generic filter $G$.
It is also known ...

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**1**answer

294 views

### Does ZF prove that proximity spaces are completely regular?

(This is based on my earlier question, but I think this one would be easier to answer.)
Let $\langle X,\mathbf{\delta} \hspace{.01 in} \rangle$ be a separated proximity space, and let ...

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728 views

### On a weak choice principle

[PLEASE SEE EDITS AT BOTTOM OF QUESTION]
Consider the following set-theoretic axiom:
For each set $X$ there exists a set-indexed collection $\{C_i \to X\}_{i\in I_X}$ of surjections such that for ...

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**1**answer

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### Axiom of choice and non measurable set

We know that existence of a Lebesgue non-measurable set is consistent with the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue non-measurable set imply the Axiom Of ...

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### Does ZF prove that topological groups are completely regular?

Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$.
Assume $\{\mathbf{e}\}$ is closed in $\langle ...

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**1**answer

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### Can a Vitali Set be constructed without AC?

For the purposes of this discussion, let a Vitali Set be any subset $V\subseteq{}[0,1)$ such that for $V_q:=\{x+q\;|\;x<1-q,\;x\in{}V\}\cup\{x+q-1\;|\;x\geq{}1-q,\;x\in{}V\}$ there is a countable ...

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### Can a Vitali set be Lebesgue measurable? (ZF)

Here is the definition of Lebesgue measure.
The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In ...

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### For models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$?

Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$.
Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if ...

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732 views

### Haar measures in Solovay's model

Haar measure is a measure on locally compact abelian groups which is invariants to translations. For example, the Lebesgue measure on the reals is such measure.
It can be shown without the use of the ...

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### 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 ...

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### On the difference between two concepts of even cardinalities: Is there a model of ZF set theory in which every infinite set can be split into pairs, but not every infinite set can be cut in half?

An interesting question has arisen over at this
math.stackexchange
question
about two concepts of even in the context of infinite
cardinalities, which are equivalent under the axiom of
choice, but ...

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votes

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### Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?

In ZFC, every construction of a Lebesgue or Borel non-measurable set uses the axiom of choice. None of them that I've seen use choice to define a unique set, even though it's entirely possible to do ...

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**1**answer

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### Axiom of choice and bases of vector spaces over a fixed field

Let $k$ be a field. In 1984 Andreas Blass proved that the axiom "for every extension $K|k$, every vector space over $K$ has a basis" implies the axiom of choice. He also raised the question
Does ...

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### Half Cantor-Bernstein Without Choice

I had a discussion with one of my teachers the other day, which boiled to the following question:
Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and ...

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### Indecomposable vector spaces and the axiom of choice

It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B)
It is also known ...

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263 views

### algebraic dual and axiom of choice

If $K$ is a field, the dual of $K^{({\mathbb N})}$ is $K^{\mathbb N}$, and axiom of choice implies that the natural map from $K^{({\mathbb N})}$ to the dual of $K^{\mathbb N}$ is
far from being ...

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688 views

### Chevalley's valuation extension theorem and the axiom of choice

Hello,
Do we know if the axiom of choice is needed for Chevalley's valuation/place extension theorem (i.e. the theorem that states that for every valued field and a field extension, one can extend ...

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### Reference Request: Independence of the ultrafilter lemma from ZF

I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"):
The ultrafilter lemma is independent of ZF.
ZF + the ultrafilter ...

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**1**answer

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### Are there any non-linear solutions of Cauchy's equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice?

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$
It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in ...

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**1**answer

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### How much choice do we need for regularity of product of regular spaces ?

It is usually stated that the (possibly uncountable) product
of regular topological spaces is regular.
However the only proof that I know of this fact seems to use the full axiom of choice :
See ...

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### Countable Fields with No Countable Extension

Let $\mathscr{S}$ be the set of all countable subfields of $\mathbb{C}$. Clearly, $\mathscr{S}$ is a partially ordered set under inclusion, and if $K_1\subseteq K_2 \subseteq \cdots$ is an ascending ...

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### If a result is apparently provable with AC, is actually independent of ZF?

Given the number of results that are independent of ZF. It seems that once you've found a proof of a theorem that uses the axiom of choice, the odds are that it will be independent of ZF. So my ...

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### Axiom of Choice in a weaker system

Is it known whether or not there is a consistent system of logic where two or all of the axiom of choice, well-ordering principle, and Zorn's lemma have no (known) proof of equivalence?
I was ...

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### Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice?

Let X be an infinite set.
Is it possible to show the existence of a countably infinite subset of X without using the Axiom of Choice?

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### Splitting infinite sets

There are two questions here, an explicit one, and another (more vague) one that motivates it:
I am pretty certain the following should have a negative answer, but at the moment I'm not seeing how to ...

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**1**answer

646 views

### Is “second-countable implies separable” equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for ...