10,947 reputation
33795
bio website boolesrings.org/asafk
location Israel
age 29
visits member for 4 years, 3 months
seen 35 mins ago

Born and raised in Israel. Ph.D. student in the Hebrew university in Jerusalem; studying set theory.


1h
comment A question on Gandy-Jensen system and the rudimentary functions
Noah, as a side remark, you should join us at the sans serif side and start denoting theories with \mathsf. It looks nicer on this side of the board. :-P
1h
comment First-order axiomatization of free groups
Interesting. The strongly compact cardinal enters from the Magidor-Vaananen work about Lowenheim-Skolem-Tarski numbers for second-order logic. Right?
15h
comment A question on Gandy-Jensen system and the rudimentary functions
I think it's high time we have a [fine-structure] tag. But I'd be happy if one of the guys that know more about it will write the tag wiki/excerpt. I feel insufficiently knowledgeable about the topic.
16h
revised A question on Gandy-Jensen system and the rudimentary functions
edited tags
1d
comment On $V$-decisive and weakly homogeneous forcings
Oh, the term of "fall back" seemed, in my head, to have the opposite meaning. I suppose it's a combination of not being a native speaker, the late hour and the lack of sleep last night. I'll talk about this with Yair tomorrow, and see if this ties up whatever it is we wanted regarding this topic. In the meantime, Gute Nacht as they say in German speaking parts of the globe! :-)
1d
comment On $V$-decisive and weakly homogeneous forcings
Joel, since the ground model is definable uniformly from a parameter in the ground model, the fact that the forcing is $V$-decisive is enough to allow this to go through without worrying about parameter free definability (remember that in the formulas we allow any canonical names to be referred to, and $\check V_\delta$, for the appropriate $\delta$ is certainly one of them!).
1d
comment On $V$-decisive and weakly homogeneous forcings
Thanks Joel! I think that the observation that you can assume $V[H]=V[G]$ is the key point that we missed today. We actually went through that blog post and through Grigorieff's paper, and everything seems to would have fallen into place, if only we could have assumed $V[H]=V[G]$. And indeed this is the case here!
1d
comment On $V$-decisive and weakly homogeneous forcings
@Joel: Agnostic is an interesting name, I'm not sure if it's better than $V$-decisive. But surely this is not a concept that was introduced for the first time on MathOverflow! :-)
1d
comment On $V$-decisive and weakly homogeneous forcings
@Ali: Yair insisted that this cannot be new, I said that it might be. I originally suggested "pseudo-homogeneous" and today Yair came up with this name instead, which I also think is a reasonable name.
1d
comment On $V$-decisive and weakly homogeneous forcings
@Ali, yes that is essentially the negation of a counterexample. Only stronger and quite a natural follow up. Do you happen to know the answer to the side question about the name?
1d
comment On $V$-decisive and weakly homogeneous forcings
@Ali: Yes, that was our essential question, and in light of Joel's answer we in fact noticed a mistake in the question (well, two of them really). Now it should be better, I hope. :-)
1d
comment On $V$-decisive and weakly homogeneous forcings
It seems that your answer was cut short (and the LaTeX still needs some work). In either case, the meaning was that the Boolean completion of the forcing is not weakly homogeneous, and in your example this is not the case, since the Boolean completion is just the lottery sum of $\kappa$ copies again. In either case, this is not quite what we were looking for, and I've edited the question to clarify (and remove the mistake, it shouldn't be dense embedding, just an embedding). Thank you for your help clarifying the question, and for the answer so far! Our deepest apologies for the extra work. :)
1d
revised On $V$-decisive and weakly homogeneous forcings
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1d
comment On $V$-decisive and weakly homogeneous forcings
But we (myself and Yair, that is) asked about rigid completions, not rigid posets. Also, note that we didn't mention isomorphisms between cones, but rather embeddings. We already came up with an example (which was known, apparently) of a rigid partial order whose completion is the same as the collapse of a certain cardinal to $\omega$. So the question is fundamentally about the Boolean completions. I'm sorry if it wasn't clear enough from the question! (But thanks for the answer, it's always nice to hear that we didn't make mistakes in our own arguments before! :-))
1d
comment On $V$-decisive and weakly homogeneous forcings
Yes, of course. Thank you.
1d
revised On $V$-decisive and weakly homogeneous forcings
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1d
asked On $V$-decisive and weakly homogeneous forcings
1d
comment Proofs of the uncountability of the reals.
@Joel: Does the proof using $\aleph_0$-categoricity of DLO use diagonalization somewhere? (Just curious to hear your opinion.)
Oct
17
comment Proofs of the uncountability of the reals.
This is exactly the argument of nested intervals. By the way, the fact that Bob might not have a winning strategy doesn't mean that Bob has to lose.
Oct
17
comment Proofs of the uncountability of the reals.
@Andres: Fine, so if $2^\omega\leq\Bbb R$, why do we have to work anymore than that? This suffices to show that $R$ is uncountable, since there is no surjection from $\omega$ onto $2^\omega$.