2
votes
1answer
38 views

What is the definition of the Feferman-Levy model?

Any (reference to) definition of Feferman-Levy model in set theory? I cannot find any... Though I know what is Levy collapse.
5
votes
1answer
70 views

Partial order on cardinalities without the axiom of choice

Cardinality can still be defined without choice, e.g. as equivalence class of equipotent sets, see Defining cardinality in the absence of choice. Injections define partial order on cardinalities by ...
4
votes
0answers
46 views

Strength of “Every finite dimensional subspace of a vector space has a complement”

Does the following choice principle have a name? Every finite dimensional subspace of a vector space has a complement. Equivalently, every line inside a vector space has a complementary ...
3
votes
1answer
67 views

Book/Books leading up to the the axiom of choice?

I am familiar with the axioms of ZF set theory and some basic uses of them to completely formally construct more complex objects such as natural numbers etc. However I have pretty much no background ...
3
votes
1answer
58 views

existence of spanning trees in complete graphs implies choice?

it is known that the existence of spanning trees in arbitrary (connected) graphs implies the Axiom of Choice. I was wondering if this result still holds if we restrict ourselves to spanning trees of ...
4
votes
2answers
123 views

The regularity of successor cardinal

I was looking at two different proofs of the fact that successor cardinals are regular. It struck me as odd that both proofs used AC. Looking at the concepts involved in defining cofinality I feel as ...
2
votes
1answer
30 views

Existence of transversals of subgroups implies axiom of choice?

If $G$ is a group and $H\leq G$ is a subgroup, then a transversal of $H$ is a subset $T\subseteq G$ which meets every coset of $H$ in a unique point. The axiom of choice clearly implies that every ...
2
votes
4answers
131 views

Boolean prime ideal theorem and the axiom of choice

The Boolean prime ideal theorem is strictly stronger than ZF, and strictly weaker than ZFC. I'm looking for nice examples (like the existence of non-measurable set) that request at least that theorem ...
7
votes
0answers
259 views

Proving equivalence of a tree-based version of Countable Choice for families of finite sets.

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
3
votes
1answer
107 views

Dependent choice does not imply “the reals are well-ordered”; citation?

As silly as this sounds, I can't find a proof that the axiom of dependent choice (DC) does not imply that the reals are well-orderable. My memory is that this is a fairly early result in the history ...
1
vote
2answers
98 views

Is there a deductive system for second-order logic that is complete with respect to Henkin semantics?

I have heard that second-order logic with Henkin semantics is a lot like first-order logic. Does this mean it has a complete deductive system? If so, what's an example of such a deductive system, and ...
7
votes
0answers
223 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 know that Choice principles have some connection to the separation axioms (in ZF, at ...
2
votes
1answer
181 views

How do we know we need the axiom of choice for some theorem?

I have been working through Munkres Topology book and in an exercise he says that there was a theorem he proved in a previous section that relied on the axiom of choice and the task is to find it. I ...
5
votes
1answer
173 views

In ZF, does there exist an ordinal of provably uncountable cofinality?

Question is in the title. In ZFC, one can prove that $\aleph_{\alpha+1}$ is regular, so there is a large source of cardinals with uncountable cofinality, but in ZF, it is consistent that ${\rm ...
5
votes
2answers
105 views

Reference request, self study

I'm looking for references (books/lecture notes) for : Cardinality without choice, Scott's trick; Cardinal arithmetic without choice. Any suggestions ? Thanks in advance.
6
votes
3answers
198 views

Questions on Fraenkel models

Halbeisen on page 172 contains a section entitled "The Second Fraenkel Model". The original paper by Fraenkel containing this model can be found here. I have several questions regarding this model and ...
4
votes
2answers
107 views

Are there versions of the axiom of choice that restrict the size of the factors?

One formulation of the axiom of choice is that an arbitrary product of nonempty sets must be nonempty. The axiom of countable choice AC$_\omega$ is known to be strictly weaker than AC, but still ...
9
votes
1answer
100 views

Proof of a basic $AC_\omega$ equivalence

On Wikipedia it is mentioned that "... in order to prove that every accumulation point $x$ of a set $S\subseteq \mathbf R$ is the limit of some sequence of elements of $S\setminus \{x\}$, one uses (a ...
4
votes
1answer
169 views

Books on Axiom of Dependent Choices?

Are there any books about the axiom of dependent choice? There are books about the axiom of choice, e.g. Herrlich or Jech. But I can't seem to find any on the axiom of dependent choices. As far as I ...
3
votes
2answers
115 views

Cohen and the axiom of choice

The wikipedia article on Paul Cohen mentions that: Cohen is noted for developing a mathematical technique called forcing, which he used to prove that neither the continuum hypothesis (CH), nor ...
10
votes
1answer
678 views

Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly

While reading David Williams's "Probability with Martingales", the following statement caught my fancy: Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel ...
33
votes
2answers
2k views

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ ...
22
votes
5answers
781 views

Axiom of choice - to use or not to use

I was wondering if there are examples of results in mathematics that were first proven using axiom of choice and later someone found a proof of the result without using the axiom of choice.