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1h
comment The sequence $x_{n+1}=ax_{n}+b $ converges to where?
A nitpick: Your rule only works for $x_0=1$, otherwise it's $x_0 a^n$ instead of $a^n$.
1d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Yeah, I meant the second version with the constant symbol, sorry I wasn't being clear enough on that. Ah but I don't think "that model itself is ill-founded" is an issue, because that's just talking about being externally illfounded, in contrast the axiom of foundation holds inside that model, so just for sake of consistency of the theory, that should be irrelevant. (We might as well take that overarching model to be V. )
1d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Sure there are such models that contain no transitive models. But saying that there exists a model of ZFC with a transitive element that is a model is different to saying that every model of ZFC contains a model like that. Showing that every finite subset of a theory is consistent is by the compactness theorem enough to show that the theory is consistent. In this case, the theory would be "ZFC+(there exists a transitive model of ZFC)".
2d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
But we can show from the consitency of ZFC the consistency of ZFC+there is a transitive model: Given a finite subset of ZFC, by reflection in any model of ZFC there is a $\theta$ so that $V_{\theta}$ satisfies this subset, so ZFC+(there is transitive M with $M\models$(this subset)) is consistent.
2d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Actually, ZFC seems to be equiconsistent with ZFC+there is an enbedding of transitive set models with the critical point not being inaccessible in the first.
2d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Right. At least we could see by that argument that with reflection, consistency of there is a transitive model of ZFC implies consistency of there is an uncountable transitive model of ZFC without inaccessibles, hence of there is an embedding with critical point which is not inaccessible. But that's not as good.
2d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Ah sure time travel and Hebrew, I will study that next. :D For now, I'm learning Spanish and set theory.
2d
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Oh right! I didn't notice the hole either, thinking of the argument to prove diamond in L with the elementary substructure of $L_{\omega_1}$. Couldn't we argue like this: Take the least $\kappa$ with $M\models \, \kappa$ is inaccessible. Then in $V_{\kappa}^M$ ZFC+"there is no inaccessible" holds. The compactness theorem + transitivation should imply there is an uncountable transitive model of ZFC+no inaccessibles (argue with constant symbols $c_{\alpha}$ for $\alpha<\omega_1$ and sentences $c_{\alpha}\neq c_{\beta}$ for $\alpha <\beta<\omega_1$).
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
In understand, but I didn't know yet if the most general form of question has an example, although I thought there is probably one, I didn't know how difficult it is, or if restrictions wil make it harder to answer. I also thought about asking : What if $M$ is a model of ZFC minus powerset and $\kappa$ is the largest cardinal in $M$. Anyway, I got some nice answers, so I'm happy.
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Ok, I see. Then I'll be able to go from there. Thanks again for the swift response. :) It seems like there is a science about this kind of forcing. A course on large cardinals in hebrew, that's pretty specific. :)
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Thanks all the same.
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Thank you. I'll have to read into these 3 (with Camilo's) examples. It'll be fun I guess. About the first example, which forcing do you mean when you say "we can force over M"? Or is that better left as an exercise to figure out?
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
Ah, I knew this part and I have read about amenable ultrafilters in the context of iterated ultrapowers. I looked in "The Higher Infinite" by Kanamori again and he points out you only need models of ZFC without the powerset axiom to define the M-ultrafilter. (in case M and N have the same powerset of $\kappa$)
Aug
31
comment Is the critical point of an embedding of a model of set theory inaccessible in it?
This is the simplest example to me. Thank you.
Aug
30
asked Is the critical point of an embedding of a model of set theory inaccessible in it?
Aug
30
comment What's wrong with this proof of Schröder-Bernstein theorem?
The argument works, it is the same that can be found for example in Jech's "Set Theory". I think "analyze" just means checking it is correct here.
Aug
30
comment Closeness of measures on a cardinal
I checked that for a two-step iteration by a normal measure, $V\to M_1 \to M_2$, the $k$ obtained from the map from $V$ to $M_2$ is the same as the map from $M_1$ to $M_2$. And the embedding of $V$ to $M_2$ is an ultrapower embedding for a product ultrapower on $\kappa\times \kappa$, but we could make that into an ultrapower on $\kappa$ with Gödel pairing. So in this case the critical point of $k$ is $j_1(\kappa)$ when $j_1:V\to M_1$ is the ultrapower embedding for the normal measure on $\kappa$.
Aug
29
revised Looking for extender axioms
added 407 characters in body
Aug
29
comment Looking for extender axioms
Haha, well English is not my native language, maybe the title sounds weird.
Aug
29
revised Looking for extender axioms
edited tags