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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.

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 Cohen-subset 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 Cohen-subset 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 Cohen-subset 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 well-order 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 classes-of-classes-of-classes 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 non-transitive 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
Jan
29
revised Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?
added 82 characters in body
Jan
29
asked Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?
Jan
11
awarded  Nice Question
Jan
10
accepted Is every class that does not add sets necessarily added by forcing?
Jan
10
comment Is every class that does not add sets necessarily added by forcing?
What a great example, Joel! The fact that the satisfaction class is compatible with GBC but is not definable is a very nice combination. Does existence of a satisfaction class increase the consistency strength of GBC?