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14
votes
1answer
347 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 ...
7
votes
2answers
237 views

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 ...
1
vote
1answer
99 views

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 ...
10
votes
2answers
261 views

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 ...
2
votes
1answer
127 views

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 ...
5
votes
2answers
154 views

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 ...
9
votes
0answers
173 views

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 ...
8
votes
1answer
451 views

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 ...
5
votes
1answer
202 views

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 ...
22
votes
15answers
1k views

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 ...
4
votes
1answer
183 views

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 ...
9
votes
1answer
298 views

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 ...
8
votes
1answer
378 views

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 ...
21
votes
2answers
1k views

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 ...
2
votes
0answers
294 views

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 ...
3
votes
1answer
344 views

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 ...
13
votes
1answer
826 views

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 ...
36
votes
1answer
1k views

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 ...
4
votes
2answers
612 views

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 ...
3
votes
3answers
453 views

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 ...
11
votes
3answers
1k views

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 ...
8
votes
3answers
776 views

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 ...
2
votes
0answers
259 views

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 ...
2
votes
2answers
548 views

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 ...
7
votes
2answers
605 views

Category and the axiom of choice

What are (if any) equivalent forms of AC (The Axiom of Choice) in Category Theory ?
2
votes
1answer
344 views

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 ...
12
votes
4answers
888 views

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?
0
votes
2answers
241 views

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 !? ...
2
votes
2answers
272 views

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 ?
5
votes
1answer
293 views

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 ...
4
votes
5answers
1k views

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 ?
2
votes
3answers
261 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
231 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
226 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
287 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
708 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
654 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 ...
12
votes
0answers
377 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
212 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
725 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 ...
7
votes
1answer
350 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
341 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
351 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
303 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
329 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
498 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
427 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
358 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
400 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 ...