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Apr
7
awarded  Popular Question
Dec
27
comment Looking for counter example - compactness theorem
I say that it doesn't even matter that $S$ is a collection of sets. It could've been any set, and your $x_A$ is a function $f:S\to\mathbb{N}$ (or what ever range you are required), that should satisfy $f(A)+f(B)+f(C)\neq 0 \mod 10$. Whether such a function exists or not does not depend on the nature of the elements of $S$.
Dec
27
comment Looking for counter example - compactness theorem
It seems to me that the fact that $S$ is a family of sets doesn't even matter. The fact that $A,B,C\in S'$ is the only property you need. I can't see how you could even use the finite sets assumption.
Dec
19
answered extend a linear function
Dec
17
awarded  Caucus
Jul
12
awarded  Good Question
Mar
29
awarded  Announcer
Feb
4
comment $SAS^{-1}=\lambda A$ - show $\lambda^n=1$or A is nilpotent
It is. What you said is obviously equivalent to $|\lambda|=1$, which I showed in the first step of my proof. @Sami Ben Romdhane showed that too.
Feb
3
answered $SAS^{-1}=\lambda A$ - show $\lambda^n=1$or A is nilpotent
Feb
3
comment $SAS^{-1}=\lambda A$ - show $\lambda^n=1$or A is nilpotent
This proof is wrong, since $\mu, \lambda\mu, \ldots, \lambda^n\mu$ aren't necessarily distinct. Actually, this question is wrong, I'll post my answer in a few minutes.
Jan
8
accepted Embedding models of ZF into another model
Jan
4
comment Embedding models of ZF into another model
Using Löwenheim–Skolem we can also find a model which is of minimal cardinality (which will be $\max\{|M|,|N|\}$). Do you have any idea regarding finding a minimal model with respect to embedding?
Jan
4
comment Embedding models of ZF into another model
This is interesting, I didn't think of using compactness. This indeed proves that we can find a model in which both models can be embedded, so this answers the first question. The same argument applies for an arbitrary large collection of models.
Jan
4
comment Embedding models of ZF into another model
I understand, thank you again for your time :)
Jan
4
comment Embedding models of ZF into another model
Thanks! This is exactly the sort of answers I was looking for. I will go through it later on. I am still trying to understand why did this question get the down vote...
Jan
4
comment Embedding models of ZF into another model
That is about what I thought about. But the dijoint union is not enough, since the axiom of pairing wouldn't hold. What I figured out that this can be defined categorically as the dijoint union in the category of models of ZF with homomorphisms as the morphisms.
Jan
4
comment Embedding models of ZF into another model
Maybe this is the answer I was looking for. I was thinking that even if two models aren't elementarily equivalent, we can still amalgamate (new word for me btw) them. For example if in a model AC does not hold maybe we can "add enough sets" to "fix" that. Maybe I was wrong...
Jan
4
comment Embedding models of ZF into another model
@hot_queen, thank you, however I am not asking about elementarily equivalent models, but any two models.
Jan
4
comment Embedding models of ZF into another model
I agree that this question is very broad, but this what I am looking for, interesting results that incorporate this idea. I don't really care about the exact details, but about the concept.
Jan
4
asked Embedding models of ZF into another model