All Questions
Tagged with model-theory filters
60
questions
3
votes
0
answers
62
views
Two families of isomorphic structures have isomorphic ultraproduct.
I am trying to prove the following result:
Let $(\underline{M}_i)_{i\in I}$, $(\underline{N}_i)_{i\in I}$ be two families of structures such that, for all $i\in I$, $\underline{M}_i \cong \underline{...
- 103
4
votes
0
answers
115
views
What is the power of a set
I'm currently reading the article Ultraproducts and saturated models by H.J. Keisler (link: https://core.ac.uk/download/pdf/82090773.pdf). In lemma 1.1, he talks about a set having power $\alpha$. I'm ...
1
vote
0
answers
42
views
Is this a non-standard extension?
I started reading Henson's "Foundations of Nonstandard Analysis. A Gentle Introduction to Nonstandard Extensions" a couple of days ago, and I am a bit confused about something.
Let $F$ be an ...
- 3,545
2
votes
1
answer
102
views
Measurable Cardinals, Elementary Embeddings, and building up to transitive collapse
I am looking at the proof that $\kappa$ a cardinal is measurable iff it is the critical point of an elementary embedding from $V$ to an inner model $M$, specifically the forward direction.
I ...
- 95
3
votes
2
answers
163
views
What is the relationship between ultrafilters and propositional theories?
I am learning more about how to use ultrafilters by using them to prove several of the typical results which appear as applications of propositional (or first-order) compactness.
Generally speaking, ...
- 4,432
6
votes
2
answers
293
views
Filter/Ultrafilter Methods in Commutative Algebra/ ring theory
I have a rather broad question: Are there any useful results/ approaches known on study commutative unitary rings and their ideals with filter/ ultrafilter methods?
What I know:
In Boolean algebra ...
- 7,571
2
votes
2
answers
182
views
Is the cardinality of an ultrapower generally the same as that of the set of functions?
For any structure $M$ and ultrafilter $U$ on a set $\Omega$, let $M^{\langle U\rangle}=M^\Omega/U$ be the conventional ultrapower on $M$.
Is it the case that $|M^{\langle U\rangle}|=|M^\Omega|=|M|^{|\...
- 5,040
0
votes
1
answer
199
views
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
6
votes
0
answers
100
views
In ZFC, does every strucutre have an ultrapower that is universal?
I call a structure $M$ of cardinality $\kappa$ universal if it is $\kappa^+$-universal, i.e., there is an elementary embedding from any $N \equiv M$ to $M$ if $|N| \le \kappa$. (I think this is a ...
- 7,290
1
vote
0
answers
112
views
Existence of ultrapower of arbitrary cardinality
Given an infinite structure $\mathcal M$ and a cardinal $\kappa$, prove there exists an ultrapower $\mathcal M^I/U$ that has cardinality $\kappa$.
The motivation is to use ultrapower to prove upwards ...
- 361
0
votes
0
answers
54
views
Computing the size of a definable set in a model from its size in its ultrapower
Suppose $\lambda$ is a cardinal and $U$ is an regular ultrafilter on $\lambda$. For a formula $\phi$ and a structure $M$, if $\phi(M)$ is infinite, then $|\phi(M^\lambda /U)| = |\phi(M)|^\lambda$. ...
- 7,290
9
votes
1
answer
317
views
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
5
votes
0
answers
103
views
Does there exist a $\lambda$-good, $\omega$-incomplete filter on $\kappa$ such that $P(\kappa)/\lambda$ is $\lambda$-saturated?
Let $\kappa$ and $\lambda$ be cardinals (we can assume $\lambda < 2^\kappa$). Does there exist a $\lambda$-good, $\omega$-incomplete filter on $\kappa$ such that the quotient $P(\kappa)/\lambda$, ...
- 7,290
3
votes
1
answer
251
views
Measurable cardinals, elementary embeddings, and Kunen's theorem
Suppose $\kappa$ is a measurable cardinal. Then if $U$ is the ultrafilter on $\kappa$, we can use this to generate an ultrapower of the entire universe. We can then embed $V$ into this ultrapower in ...
- 8,000
1
vote
1
answer
66
views
Concise argument: $[f]$ in the ultrafilter that realizes $\Sigma$?
Let $\mathcal{L}$ be a countable language and $\mathcal{A}_i$ be $\mathcal{L}$-structures.
Let $\mathcal{A}=\prod_{i \in \omega} \mathcal{A}_i/\mathcal{U}$, where $\mathcal{U}$ is a non-principal ...
- 1,847