All Questions
Tagged with axiom-of-choice model-theory
42
questions
0
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2
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95
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How do we reconcile Solovay’s Model with the Hanh Banach Theorem in ZF
From Wikipedia, I have the following two facts:
Solovay’s Model is a model of ZF in which all sets are measurable. In this way Solovay showed that in the proof of the existence of a non-measurable ...
- 33
0
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0
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41
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Are cardinals always partial-well-ordered without AC? [duplicate]
With AC, we have that cardinals are well-ordered. Without AC, we can have two sets which are incomparable, meaning that, at best, cardinals are only partially ordered.
But do we still have, without AC,...
- 8,000
2
votes
1
answer
64
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non-wellordered linear order that doesn't contain a copy of $\omega^*$ in ZF?
Obviously a wellorder cannot contain a copy of $\omega^*$ (the dual order of $\omega$), and this can be proven in ZF. In ZFC, it is easy to prove any linear order which is not a wellorder does contain ...
- 5,633
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1
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199
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In what sense, precisely, are free ultrafilters "not definable"?
Note: The following attempted summary consists of follow-up questions to this answer by Asaf Karagila to a related question on MathOverflow. These questions are too long for a comment on that answer, ...
- 1,458
0
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1
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61
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$ZF + AC_\omega$ prove the propositional compactness theorem for arbitrary languages
Let $ZF$ be the Zermelo-Fraenkel set theory and let $AC_\omega$ be the countable axiom of choice.
In $ZF+AC\omega$ one can prove the propositional compactness theorem for countable languages.
My ...
- 455
9
votes
1
answer
317
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Can ultrapowering add choice?
I'm supervising a reading course in set theory, and the following question came up (let $\mathsf{LT}$ be Łoś's Theorem; in an earlier version of this question I mistakenly thought $\mathsf{LT}$ was ...
- 249k
8
votes
1
answer
189
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How hard is it to show $\mathsf{ZF-Pow}\not\vdash\mathsf{AC}$?
This is motivated by comments on this question.
Let $T_0=\mathsf{ZF-Pow}$ and $T_1=\mathsf{ZF-Pow+\neg AC}$. (Note that there is some subtlety about what precisely $\mathsf{ZF-Pow}$ is; here, I adopt ...
- 249k
1
vote
1
answer
68
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Function $f: \omega \to A$ with $f(0)=a$ and $f(i)Rf(i+1)$ in ZF
Let $\mathcal{L}$ be a language containing a binary relation symbol $P$. Let $A$ be a set such that, for any $\mathcal{L}$-structure $\mathcal{A}$ having domain $A$ and any $a \in A$, there exists a ...
- 1,847
4
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2
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358
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What is an example of model of ZF with a countable family of pairs whose union is uncountable?
I was reading the book "Howard and Rubin, “Consequences of the Axiom of Choice”.
For example, why do not exists sets that are neither countable nor uncountable?
It is interesting to point out ...
- 87
1
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1
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41
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An example of a class of structures whose axiomatizability depends on the axiom of choice
Is there an example of a specific class $K$ of structures (in the sense of first-order logic), such that $K$ is first-order axiomatizable if the axiom of choice is assumed, but that its ...
- 21.5k
4
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4
answers
233
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Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom?
Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$
Is $\text{ZF}\vdash \exists x\varphi(x)$ true?
- 1,247
7
votes
1
answer
268
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Are these "finite-ish" sets closed under union?
Now asked at MO.
Throughout, we work in $\mathsf{ZF}$.
Say that a set $X$ is $\Pi^1_1$-pseudofinite if for every first-order sentence $\varphi$, if $\varphi$ has a model with underlying set $X$ then $\...
- 249k
5
votes
1
answer
355
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The Upwards and Downwards Lowenheim Skolem Theorem together imply the axiom of choice (in ZF)
Here the hints were to use the facts that $ZF \vdash (\forall$ infinite$ A |A \times A| = |A|) \rightarrow AC$ and that there is a bijection between $\omega \times \omega$ and $\omega$.
My idea was ...
3
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1
answer
112
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In the foundations of NSA with ultrapowers, how much can the axiom of choice be weakened?
The derivations I've seen of the hyperreals using ultrapowers use the axiom of choice and Zorn's lemma a lot. But looking closer, you can possibly weaken the axioms used in the derivations of some ...
- 649
6
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1
answer
134
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Finiteness notions induced by $\forall\exists$-sentences
Throughout, work in $\mathsf{ZF}$ and restrict attention to finite languages.
Let $\mathsf{Ded}$ and $\mathsf{Amo}$ be the classes of Dedekind-finite and amorphous sets respectively. For $T$ a ...
- 249k