User ant emyy lee - MathOverflow

Finding questions between functional analysis and set theory

2010-03-22T14:05:20Z

Are there some good questions on functional analysis whose solution depends on tools in set theory? My major is mathematical logic, I think tools in set theory, especially infinity combinatorics and forcing, should be used to solve some questions in functional analysis. For functional analysis, I just have read the main part of Conway's textbook. In this book, I have not found such questions.

a question about diagonal prikry forcing

2010-02-10T09:39:26Z

Suppose <\kappa_n|n<\omega> is a strictly increasing sequence of measurable cardinals,

\kappa is the limit of this sequence. For each n<\omega, U_n is a normal measure on

\kappa_n. P is the diagonal Prikry forcing corresponding to \kappa_n's and U_n's. Suppose g is P-generic sequence over V. We have known that for each strictly increasing

sequence x of length \omega such that each x(i)<\kappa_i and x\in{V}, x is eventually

dominated by g. In V[g], suppose A is a subset of \kappa, A is not in V. Is there a strictly

increasing sequence y of length \omega such that each y(i)<\kappa_i and y\in{V[A]}, y is not

eventually dominated by g?

(g can eventually dominate all such sequences in V, V[A] is greater than V, I feel g can not

eventually dominate all such sequences in V[A].)

On the independence of the Kurepa Hypothesis

2011-06-22T16:38:52Z

Kurepa Hypothesis says there is a Kurepa tree, which is a $\omega_1$-tree has at least $\omega_2$ many branches. It is known that beginning from a model with an inaccessible cardinal $\kappa$, after collapes $\kappa$ to $\omega_2$ using the Levy collape, then in the generic extension, Kurepa Hypothesis fails. In above generic extension, $\omega_2$ is equal to $\kappa$ and by a counting argument for nice names, $2^{\omega_1}=\omega_2$. My question is that "is it consistent that Kurepa Hypothesis fails and $2^{\omega_1}>\omega_2$?" (The reason I think this question: The biggest possible value of the number of branches is $2^{\omega_1}$, so in the environment of $2^{\omega_1}=\omega_2$, it is most difficult for the living of a Kurepa tree. So I want to know whether this requiement is necessary.)

A question on ultrapower

2010-02-02T02:13:05Z

Suppose $\kappa_0$ is a measurable cardinal and $\mu_0$ is a normal measure on $\kappa_0$. $M_1$ is the transitive collapse of $Ult(V,\mu_0)$, $j_{0,1}:V\rightarrow{M_1}$ is the elementary embedding induced by the ultrapower. In $M_1$, $\kappa_1=j_{0,1}(\kappa_0)$ is a measurable cardinal and $\mu_1$ is a normal measure on $\kappa_1$ in $M_1$ such that $\mu_1$ is not in the range of $j_{0,1}$. $M_2$ is the transitive collapse of $Ult(M_1,\mu_1)$, $j_{1,2}:M_1\rightarrow{M_2}$ is the elementary embedding induced by the ultrapower. $j_{0,2}=j_{1,2}\circ{j_{0,1}}$.

Is it true that: ``Suppose $N$ is an inner model, $i:V\rightarrow{N}$ and $k:N\rightarrow{M_2}$ are elementary embeddings such that $k\circ{i}=j_{0,2}$. Then $k''N=j_{0,2}''V$ or $k''N=j_{1,2}''M_1$ or $k''N=M_2$''?

Comment by Ant emyy Lee

2011-06-24T14:12:32Z

I find a paper: Random trees under CH James Hirschorn Israel Journal of Mathematics, 2007, Volume 157, Number 1, Pages 123-153 I just read its abstract, he says there is a model of $\neg{KH}$, and we can add many random reals and preserve $\neg{KH}$. But I think $Add(\omega,\omega_1)$ does not add Kurepa tree over Silver model is also a question.

Comment by Ant emyy Lee

2011-06-24T05:00:14Z

I see it.Thanks.

Comment by Ant emyy Lee

2011-06-23T07:32:02Z

Thank you for your answer, Joel. Another thing is unknown for me: In Silver's model, no new real has been added, and inaccessible $\kappa$ is collaped to $\omega_2$, so Continumm hypothesis holds in this model. But if adding some Cohen reals to this model, maybe some new Kurepa tree are created. This should be my second question: is it consistent that "Kurepa hypothesis fails and $2^{\omega_1}>\omega_2$ and $2^{\omega}>\omega_1$"?

Comment by Ant emyy Lee

2011-06-23T05:10:45Z

sorry,I do not know what is community wiki. So I just choose it.

Comment by Ant emyy Lee

2010-03-22T14:48:01Z

Thank you for your solution. Axiom of choice or transfinite induction, I think, they are general tools in math, but not just in logic.

Comment by Ant emyy Lee

2010-03-22T14:44:49Z

This book should be what I was searching for. Thank you.

Comment by Ant emyy Lee

2010-03-22T14:32:24Z

en. This is a good example. Also, do you know any questions as this style which is still open now?

Comment by Ant emyy Lee

2010-02-11T06:53:36Z

Thank you for your astonishing solution.

Comment by Ant emyy Lee

2010-02-11T02:52:12Z

This forcing has Prikry property. Similarly to Prikry forcing, if \kappa is the limit of this measurable cardinal, then this forcing does not add any new bounded subset of \kappa. This forcing appears in the chapter "Prikry-type forcing" of Handbook of set theory written by Moti Gitik. It is in the section 1.3 of this chapter.

Comment by Ant emyy Lee

2010-02-11T02:47:49Z

yes. I mean for every A, such that A\subseteq{\kappa} and A\notin{V}, there is such a sequence in V[A] not dominated by g.

Comment by Ant emyy Lee

2010-02-11T02:43:17Z

Maybe I did not say clearly. You find an A, there is a sequence in V[A], g can not eventually dominate it. But I want to know whether for every A such that A\subseteq{\kappa} and A\notin{V}, there is such a sequence in V[A] not dominated by g.

Comment by Ant emyy Lee

2010-02-10T16:50:17Z

Hi. Every condition of P is a ordered pair (s,F). s is a strictly increasing finite sequence such that each s(i)<\kappa_i. F is a function, dom(F)=\omega, for each i<\omega, F(i)\in{U_i}. (s,F) and (t,H) are two conditions. (s,F) is stronger than (t,H) means: (i) s end extends t; (ii) for each i<\omega, F(i) is a subset of H(i); (iii) for each i, if |t|\leqslant{i}<|s|, s(i)\in{H(i)}. Also, I do not know whether this forcing should be called ``diagonal prikry forcing''.

Comment by Ant emyy Lee

2010-02-03T03:20:28Z

So the model-theoretic fact you said in your answer may not make sense in this case, since $\mu_1$ is not a normal measure in $V$. How to deal with this case?

Comment by Ant emyy Lee

2010-02-03T03:20:13Z

I am the student exchange you with email, you introduce me to go here, so I register this new name. In my question, I require $\mu_1$ is not in the range of $j_{0,1}$, i.e. the second step of the iterated ultrapower is done by a normal measure not in $j_{0,1}''V$. For example, suppose $\kappa$ has $\kappa$ many normal measures <U_\alpha:\alpha<\kappa>, the first step of the iterated ultrapower I use $U_0$, and the second step of the iteration, I use the \kappa-th measure in the sequence j_{0,1}(<U_\alpha:\alpha<\kappa>) .