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17
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1answer
1k 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
526 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
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0answers
270 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
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0answers
549 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
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0answers
332 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
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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 ...
17
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6answers
1k views

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
753 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 ...
22
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1answer
1k views

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
votes
0answers
775 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 ...
5
votes
2answers
487 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 ...
5
votes
2answers
511 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 ...
11
votes
1answer
574 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 ...
3
votes
1answer
232 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 ...
5
votes
2answers
658 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 ...
12
votes
4answers
974 views

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 ...
15
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1answer
2k views

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 ...
6
votes
1answer
384 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 ...
4
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1answer
289 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 ...
9
votes
3answers
721 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 ...
3
votes
1answer
1k views

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 ...
16
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1answer
718 views

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 ...
2
votes
1answer
359 views

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 ...
18
votes
2answers
2k views

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 ...
7
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1answer
333 views

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 ...
16
votes
2answers
720 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 ...
21
votes
2answers
1k views

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 ...
29
votes
2answers
1k views

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 ...
6
votes
5answers
904 views

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 ...
15
votes
1answer
979 views

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 ...
18
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3answers
2k views

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 ...
11
votes
2answers
784 views

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 ...
2
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0answers
260 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 ...
4
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3answers
666 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 ...
6
votes
3answers
686 views

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 ...
10
votes
1answer
2k views

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 ...
3
votes
1answer
401 views

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 ...
2
votes
2answers
736 views

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 ...
1
vote
4answers
827 views

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 ...
8
votes
1answer
593 views

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 ...
6
votes
3answers
2k views

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?
20
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1answer
1k views

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 ...
2
votes
1answer
623 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 ...
20
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0answers
976 views

Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property

While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then ...
7
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3answers
727 views

Construction of a maximal ideal

Hello, Let R denote the ring of continuous functions defined on the real line, let I in R be the ideal consisting of functions with compact support. Obviously, I is not maximal, and by Zorn's Lemma ...
2
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1answer
405 views

Nets and the Axiom of Choice

Suppose that $f : X \rightarrow Y$ is mapping between topological spaces that is not continuous at $x_0$. Then there is an open set $V$ in $Y$ containing $f(x_0)$ such that for any open set $U$ ...
10
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1answer
660 views

Dynamical systems, minimal sets and the Axiom of Choice

Perhaps the most important application of the Axiom of Choice within the theory of dynamical systems (meaning here, compact Hausdorff spaces with a self-map) yields, within every dynamical system, the ...
33
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6answers
2k views

Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The ...
36
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4answers
4k views

Does the fact that this vector space is not isomorphic to its double-dual require choice?

Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their ...
9
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2answers
395 views

Completeness of the club filter without AC

Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the ...