An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty ...

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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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555 views

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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1answer
501 views

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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4answers
724 views

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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2answers
1k views

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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452 views

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 ...
7
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1answer
270 views

$\Theta$ 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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1answer
391 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 ...
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481 views

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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2answers
205 views

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 ...
7
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1answer
173 views

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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3answers
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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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3answers
891 views

About the axiom of choice, the fundamental theorem of algebra, and real numbers

About fundamental theorem of algebra, there is a large collection different demonstrations. I ask: is there some proof that avoids AC (choice axiom)? In a general topos (with natural number object) ...
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1answer
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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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3answers
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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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1answer
629 views

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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3answers
594 views

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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190 views

Logical relationships between weakenings of AC

What are the known logical implications between weak choice principles like $DC_\kappa$", the ...
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1answer
430 views

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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0answers
890 views

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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1answer
428 views

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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863 views

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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1answer
2k views

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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277 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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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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2answers
540 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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2answers
501 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, ...
18
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1answer
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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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555 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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272 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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592 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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350 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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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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6answers
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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 ...
13
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2answers
789 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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1answer
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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 ...
9
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0answers
846 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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2answers
495 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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2answers
529 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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1answer
616 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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1answer
240 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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2answers
670 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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4answers
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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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1answer
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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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1answer
443 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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1answer
298 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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3answers
740 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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1answer
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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 ...