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2
votes
3answers
271 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
240 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
309 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
725 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
669 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
520 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
235 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
757 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
374 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
365 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 ...
14
votes
2answers
1k views

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 ...
6
votes
1answer
374 views

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 ...
9
votes
1answer
322 views

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 ...
4
votes
0answers
377 views

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 ...
8
votes
0answers
527 views

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 ...
4
votes
1answer
434 views

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. ...
1
vote
0answers
435 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 ...
12
votes
1answer
440 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 ...
10
votes
4answers
683 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 $A$ ...
17
votes
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: $\quad$ ...
6
votes
1answer
410 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
votes
1answer
247 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 ...
5
votes
1answer
365 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 ...
9
votes
1answer
423 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 ...
6
votes
2answers
198 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
votes
1answer
164 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 ...
4
votes
3answers
301 views

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 ...
4
votes
1answer
863 views

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 ...
6
votes
2answers
954 views

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 ...
10
votes
1answer
517 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 ...
1
vote
3answers
535 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, ...
3
votes
0answers
185 views

Logical relationships between weakenings of AC

What are the known logical implications between weak choice principles like $DC_\kappa$", the ...
1
vote
1answer
379 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 ...
21
votes
0answers
761 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 ...
10
votes
1answer
412 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 ...
27
votes
0answers
1k views

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 ...
13
votes
1answer
736 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 ...
4
votes
1answer
1k 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 ...
3
votes
0answers
249 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 ...
10
votes
3answers
1k views

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. ...
16
votes
4answers
1k views

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 ...
1
vote
2answers
494 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 ...
3
votes
2answers
410 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
votes
1answer
998 views

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 $A\subseteq 2^{\omega}$ ...
6
votes
3answers
523 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 ...
6
votes
0answers
269 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 ...
16
votes
0answers
548 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 ...
3
votes
0answers
331 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 ...
34
votes
1answer
3k views

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 ...