bio | website | |
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location | New York City | |
age | 40 | |
visits | member for | 4 years |
seen | yesterday | |
stats | profile views | 193 |
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? |