bio  website  jdh.hamkins.org 

location  New York City  
age  
visits  member for  4 years, 5 months 
seen  12 mins ago  
stats  profile views  42,542 
I am a professor at the City University of New York, at the College of Staten Island and the CUNY Graduate Center, living in midtown Manhattan. My main research interest lies in mathematical logic, particularly set theory, focusing on the mathematics and philosophy of the infinite. A principal concern has been the interaction of forcing and large cardinals, two central concepts in set theory. I have worked in group theory and its interaction with set theory in the automorphism tower problem, and in computability theory, particularly the infinitary theory of infinite time Turing machines. Recently, I am preoccupied with the settheoretic multiverse, engaging with the emerging field known as the philosophy of set theory.
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Formal definition of function: equality
See related mathoverflow.net/a/30397/1946 
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Generic Ultrapower as a Class
I agree with this, and in any case, having a predicate for the ground model, as Jech does, also works fine, and avoids the need for localization. Nevertheless, to my way of thinking, the real answer to the question, "How does $M[G]$ talk about $M$?" is the ground model definability theorem. 
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Generic Ultrapower as a Class
@Ashutosh, I added some references. A complete proof is found in Laver's paper, and also in Reitz's work and elsewhere. 
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Generic Ultrapower as a Class
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answered  Generic Ultrapower as a Class 
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infinitary logic and partial fixed point logic
@NoahS, the advantage of your method, even though it uses infinitely many variables, is that it works in any language (assuming $=$ is available), whereas I needed something special about the language. 
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infinitary logic and partial fixed point logic
If one objects that the language here is infinite, then we may use instead the language with one constant $0$ and a successor operation $S$, defining the class of finite models $M$ for which the least $n$ for which $S^n(0)=0$ is in $X$. 
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infinitary logic and partial fixed point logic
We could also insist that the models have only two elements, and so what we really have is a partition of the natural numbers into at most two pieces, depending on how the constants are interpreted. 
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answered  infinitary logic and partial fixed point logic 
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infinitary logic and partial fixed point logic
Are you interested only in finite models? And do you mean $L_{\infty,\omega}$ rather than $L_{\omega,\infty}$? 
Apr 11 
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Is there an intuitionistic generalized boolean algebra (of Stone)?
The topless Boolean algebras (a Boolean algebra with top element removed) arise in the characterization of the modal logic of forcing for certain forcing classes, such as in arxiv.org/abs/1207.5841. (But that use is purely classical, rather than intuistionistic.) 
Apr 11 
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Does a BCL algebra define a partial order?
You seem to have an error in (2); do you mean $y*x=0$? 
Apr 10 
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Comparing numbers $a \uparrow^b c$ and $d \uparrow^e f$
See my related question on math.SE math.stackexchange.com/questions/72646/…, concerning in effect iterations of such expressions. 
Apr 10 
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Forcing with Nontransitive Models
If $H_\theta\models\varphi(x,\tau_G)$, with $\tau\in M$ and $\tau_G\in H_\theta$, then there is an antichain of possible values of $\tau$, and for each possible $y\in H_\theta$ that it might be, we have an $x$ for which $H_\theta\models\varphi(x,y)$. Now, by mixing $\check x$ along the antichain, we find a name $\dot x$ such that $H_\theta\models\varphi(\dot x_G,\tau_G)$. By elementarity $M\prec H_\theta$, there is such a name $\dot x$ inside $M$. And so $M[G]$ has the witness $\dot x_G$, which is one of the $x$'s that we mixed. 
Apr 10 
revised 
Quantifier elimination vs decidability
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Apr 10 
answered  Quantifier elimination vs decidability 
Apr 10 
revised 
Forcing with Nontransitive Models
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Apr 10 
revised 
Forcing with Nontransitive Models
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Apr 10 
revised 
Forcing with Nontransitive Models
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Apr 10 
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Forcing with Nontransitive Models
Vika, I think your claim that "The argument that $M[G]\prec H_\theta[G]$ works only in the case that $G$ is both fully $H_\theta$generic and also $M$generic," is not actually correct, in light of the theorem in my answer. You don't actually need that $G$ is $M$generic for this conclusion. 