Shay Ben Moshe
Reputation
539
Top tag
Next privilege 1,000 Rep.
Create new tags
4 16
Impact
~6k people reached

• 0 posts edited

# 61 Actions

 Feb 9 awarded Notable Question 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 Löwenheimâ€“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.