7
votes
1answer
261 views

Forcing, cuts, and Dedekind-finite cardinalities

Tl;dr version: there are two natural classes of cuts in the nonstandard model of arithmetic consisting of the Dedekind-finite sets (if, in fact, they constitute such a model); both these classes are ...
13
votes
1answer
847 views

How much of GCH do we need to guarantee well-ordering of continuum?

It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, ...
6
votes
0answers
192 views

Does Sageev's result need an inaccessible?

In 1981, building on work by Ellentuck in 1974, Sageev showed ("A model of ZF + there exists an inaccessible, in which the Dedekind cardinals constitute a natural non-standard model of arithmetic," ...
3
votes
1answer
232 views

What axioms (other than choice) have a taming effect on the ordering of cardinalities?

Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, ...
21
votes
1answer
575 views

Is there a model of ZF set theory with a set that does not inject into the cardinals?

Question. Is there is a model of ZF set theory with a set $X$ that does not inject into the cardinals? I use the term "cardinal" here in the ZF sense, so they are not necessarily well-orderable. To ...
6
votes
1answer
213 views

Sets of cardinalities of bases without choice

For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...
14
votes
2answers
456 views

Pathological behavior of Borel sets?

Usually in set theory, Borel sets are much more nicely behaved than arbitrary sets of reals. One reason for this is Borel determinacy, which immediately yields measurability, Baireness, and the ...
5
votes
4answers
235 views

Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?

This question comes after the comments in the recent related question Sigma-complete Lindenbaum algebras?, but in its current form is sufficiently different in my opinion, and so I decided to follow ...
5
votes
1answer
143 views

Intermediate submodels which do not satisfy AC

The following is known: Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of ...
2
votes
0answers
167 views

How many Dedekind-finite sets can $\mathbb{R}$ be partitioned into?

Building off Asaf Karagila's answer to my previous question (Can $\mathbb{R}$ be partitioned into dedekind-finite sets?) on partitioning $\mathbb{R}$ into strictly Dedekind-finite sets: (1) What ...
11
votes
4answers
836 views

Weak forms of the Axiom of Choice

Let $n\geq 2$ be a natural number and consider the following: $AC(n)$: For each family $\{X_i\}_{i \in I}$ of $n$-element sets the product $\prod_{i\in I}X_i$ is non-empty. Is it known that for ...
8
votes
1answer
245 views

Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...
2
votes
1answer
561 views

Subsets of Real Numbers (Edited & Revised Version)

Question 1: Is it consistent with $\text{ZF}$ that only countable subsets of $\mathbb{R}$ are well-orderable? Question 2: Is it consistent that for some $\lambda$, $\aleph_0 < \lambda < ...
4
votes
3answers
343 views

Minimal Generalized Continuum Hypothesis & Axiom of Choice

It is well known that working in the frame of $\text{ZF}$, the Generalized Continuum Hypothesis ($\text{GCH}$) implies the Axiom of Choice ($\text{AC}$), i.e. $\text{ZF}+\text{GCH}\vdash \text{AC}$. ...
4
votes
1answer
151 views

Discontinuous representations of GL(n,C) in ZF

Discontinuous linear representations of $GL(n,\mathbb{C})$ can be obtained from the so-called "wild" (field) automorphisms of $\mathbb{C}$; but these wild automorphisms in turn require some choice to ...
6
votes
2answers
236 views

Possible Choices for Cofinality of $\aleph_n$ without Choice

$\text{ZFC}$ proves that each $\aleph_{n}$ for $n\in \omega$ is a regular cardinal. But it seems without the Axiom of Choice there are many consistent possible choices for cofinality of such ...
2
votes
1answer
207 views

What is the order type of $L$ with Godel's well ordering?

In some sense $Ord$ is a "proper class" ordinal. Unfortunately the notion of a proper class ordinal is not a straight forward generalization of the notion of "set" ordinals because the proper classes ...
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 ...
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 ...
8
votes
1answer
379 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 ...
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 ...
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 ...
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 ...
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 ?
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
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 ...
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 ...
9
votes
3answers
563 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 $A$ ...
16
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
233 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
356 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
376 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
188 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
152 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 ...
10
votes
1answer
463 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 ...
19
votes
0answers
706 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
408 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 ...
26
votes
0answers
888 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 ...