-1
votes
0answers
105 views

Is the axiom of choice constructive in the constructible universe?

Even ZF has some non-constructive elements mostly due to contradiction proofs. For example one may be able to construct a sequence of objects some of which have a given property without being able to ...
3
votes
1answer
102 views

Why is the powerset axiom more acceptable than the axiom of choice?

The key step in Zermelo's proof of the well ordering theorem is to use $\text{AC}$ to simultaneously choose the next elelment for all possible partial chains in prospective well orderings, but that ...
8
votes
1answer
153 views

Is “PA has no non-standard models” consistent with ZF?

I have seen several proofs that there exist nonstandard models of arithmetic, but they all seem to rely on the compactness theorem, which is not implied by ZF. So are there any proofs in ZF that ...
2
votes
1answer
41 views

Are there any purely semantic proofs of the compactness theorem that don't use the full axiom of choice? [duplicate]

Using Godel's completeness theorem, it can be shown that the compactness theorem is equivalent to the ultrafilter lemma. The compactness theorem can also be proven using ultraproducts and Los's ...
8
votes
1answer
137 views

Can all theorems of $\sf ZFC$ about the natural numbers be proven in $\sf ZF$?

I know a proof of Hindman's theorem that uses ultrafilters on the natural numbers, and ultimately, the axiom of choice. But the theorem itself is essentially a combinatorial property of the natural ...
2
votes
1answer
76 views

Back and forth and the axiom of choice

Is the axiom of choice a necessary condition for the application of "back and forth construction" in model theory?
5
votes
1answer
176 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 ...
8
votes
2answers
350 views

Is the compactness theorem (from mathematical logic) equivalent to the Axiom of Choice?

Or more importantly, is it independent of the axiom of choice. The compactness theorem states the given a set of sentences $T$ in a first order Language $L, T$ has a model iff every finite subset of ...
6
votes
2answers
200 views

The Axiom of Choice and definability

I've seen a lot of relations between the notion of the existence of a definable set with a given property and the necessity of AC is proving that there is a set with the property. For example: Under ...
12
votes
3answers
234 views

Axiom of Choice and Determinacy

In my set theory course we have been talking about the axiom of determinacy. One of the first things we showed was that $AD$ and $AC$ are incompatible. We later showed that $ZF+AD$ implies the ...
4
votes
1answer
118 views

Is this theory $\kappa$-categorical when $\kappa$ is infinite and not $\le \aleph_0$?

Let $\mathcal L$ be a language with only a binary predicate $E$, and let $T$ be a theory of structures in which $E$ is an equivalence relation which partitions the structure into two infinite ...
2
votes
2answers
204 views

Is the negation of Gödel's completeness theorem consistent with $ZF$ without AC?

The proof of compactness and completeness of $\mathscr{FOL}$ (with Hilbert system) used Zorn's lemma. And Zorn's lemma is equivalent to the Axiom of choice in $ZF$. So my question is can they be ...
5
votes
1answer
335 views

Does the existence of a $\mathbb{Q}$-basis for $\mathbb{R}$ imply that choice holds up to $\frak c$?

The axiom of choice is, for ZF, equivalent to the statement that every vector space has a basis. The implication of AoC by the existence of a basis for any vector space is shown in this paper. The ...