bio  website  

location  New York City  
age  40  
visits  member for  4 years, 6 months 
seen  15 hours ago  
stats  profile views  212 
I am a professor at the New York City College of Technology of CUNY. My research focus is in set theory, particularly in the theory of forcing. I completed my graduate work at CUNY's Graduate Center under Joel Hamkins, and am lucky enough to share an office with two of my mathematical siblings.
2d

awarded  Curious 
May 22 
accepted  Does $Add(\kappa,1)^L$ ever collapse cardinals? 
May 22 
asked  Does $Add(\kappa,1)^L$ ever collapse cardinals? 
May 22 
awarded  Informed 
Jan 6 
awarded  Nice Answer 
Sep 24 
awarded  Autobiographer 
Sep 12 
answered  Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member? 
Jul 23 
awarded  Nice Question 
Feb 16 
comment 
subalgebra of a simple forcing
...I guess this just shows that $V[g][X]$ and $V[X]$ have the same reals  I'm not sure it follows that $g \in V[X]$. 
Feb 16 
comment 
subalgebra of a simple forcing
Joel, doesn't your argument work even when $\alpha$ is uncountable? That is, every real in $V[g]$ will appear as a block in $X$, regardless of the size of $\mathbb{R}$. 
Feb 2 
comment 
Can a model of set theory be realized as a Cohensubset forcing extension in two different ways, with different grounds and different cardinals?
from $W[B][C]$ up to the full model $W[B][A]$, so this does not provide an $\text{Add}(\delta,1)$ ground in the way you require. I have been playing with variations of this idea to see if I can get an example. 
Feb 2 
comment 
Can a model of set theory be realized as a Cohensubset forcing extension in two different ways, with different grounds and different cardinals?
Joel, Yes, I see. This is a very interesting question  I thought the answer was 'clearly' yes, but now I'm not so certain. One more thought, which may not prove useful: The set $A$ is not $W[B]$ generic for $\text{Add}(\delta,1)^{W[B]}$, as you correctly point out, but it is $W[B]$ generic for $\text{Add}(\delta,1)^W$. This forcing will add a generic for $\text{Add}(\delta,1)^{W[B]}$, so there is a $C \in W[B][A] \setminus W[B]$ such $W[B][C]$ is an $\text{Add}(\delta,1)^{W[B]}$ extension of $W[B]$. Unfortunately, we need to do a little further forcing (quotient forcing) to get 
Feb 2 
comment 
Can a model of set theory be realized as a Cohensubset forcing extension in two different ways, with different grounds and different cardinals?
Joel, an observation: if $M=W[A]$ is already an $Add(\delta,1)$ extension for some $\delta>\kappa$, then forcing to add a subset to $\kappa$ is equivalent to forcing over $W$ with the product, and so the answer to your question is 'yes'. If we specifically disallow $M$ of this form, then we have $M[G]=N[A]$ for which there is no $W$ with $W[A][G]=M[G]=N[A]$, which is looking an awfully lot like a counterexample to directedness of grounds (I may be ignoring some subtleties here, however.) 
Feb 2 
comment 
When Is $\mathbb{L}$Rank Definable in Inner Models of $\mathbb{V} = \mathbb{L}$?
If so, then I propose that we try to force over $\mathbb{M}$ to make $<'$ have very large order type $\beta$, so large that there is, in $L_\beta$, a function witnessing the countability of $ORD^\mathbb{M}$. Forcing to make the order type of $<'$ large presents its own challenge, but I envision adding $\omega$many Cohen reals, carefully selected from appropriate levels of the $L$hierarchy to give the correct order type. 
Feb 2 
comment 
When Is $\mathbb{L}$Rank Definable in Inner Models of $\mathbb{V} = \mathbb{L}$?
Thinking about this has led me to the question: Suppose in $\mathbb{M}$ there is a class wellorder of order type $\beta$ larger than $ORD^\mathbb{M}$. Can we (in $\mathbb{M}$) carry out the $L$construction up to level $\beta$? I realize we will not be able to define classesofclassesofclasses within $\mathbb{M}$, but I wonder if we could still define 'small' objects (e.g. reals) that arise in $L_\beta$. 
Jan 29 
comment 
Can a model of $V\neq L$ contain a class giving the $L$ordering on all its sets?
@Joel, yes, I had intended $M$ to be transitive. However, I know if we allow nontransitive models we get some very interesting and strange results, as in your "Multiverse Perspective on the Axiom of Constructibility"  so I guess I'm also interested in the nontransitive case... 
Jan 29 
comment 
Can a model of $V\neq L$ contain a class giving the $L$ordering on all its sets?
Douglas, just wanted to say I love your question. In contemplating these odd models in which every element is constructible, but $V \neq L$, I am always struck by the possibility that every set is, in fact, constructible, if we are willing to continue the $L$construction far enough beyond $ORD$. This idea is addressed more rigorously by Joel Hamkins in his A Multiverse Perspective on the Axiom of Constructibility 
Jan 29 
comment 
When Is $\mathbb{L}$Rank Definable in Inner Models of $\mathbb{V} = \mathbb{L}$?
@Andres, your answer still gives a useful starting point, since it shows that usual definition of the $L$order will not suffice to define $<_L$ on all of $M$  but it's not clear whether another definition may be more successful. 
Jan 29 
awarded  Yearling 
Jan 29 
awarded  Yearling 