489 reputation
314
bio website shaybm.co.il
location Israel
age 20
visits member for 3 years, 3 months
seen Aug 29 at 7:03

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
Nov
17
awarded  Yearling
Nov
14
revised Why is $f:\mathbb{R} \to S^1, f(t)=(\cos(2\pi t), \sin(2\pi t))$ not closed?
edited body; edited title
Nov
14
comment Is every function $f : \mathbb N \to \mathbb N$ a composition $f = g\circ g$?
Nice generalization. I wonder what would this amount to categories other than Set.
Nov
14
answered Does the position of ∃ matter?
Nov
14
comment Is every function $f : \mathbb N \to \mathbb N$ a composition $f = g\circ g$?
@azimut, indeed, actually while writing this answer I used the notation $t=g(1)$, which makes it even similar to your answer :)