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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].)

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  • $\begingroup$ Can you be more explicit about P? What are the conditions and the order, etc.? $\endgroup$
    – Joel David Hamkins
    Commented Feb 10, 2010 at 13:48
  • $\begingroup$ 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''. $\endgroup$
    – Ant emyy Lee
    Commented Feb 10, 2010 at 16:50
  • $\begingroup$ Francois pointed out correctly that there is an A for which g does not dominate all functions in V[A]. But perhaps you meant to ask whether this is true for all A not in V? Could you clarify? Also, do you know that the Prikry property holds for this forcing? That is, can we decide any given statement by shrinking only the F(i) and not exending the stem s? $\endgroup$
    – Joel David Hamkins
    Commented Feb 10, 2010 at 20:13
  • $\begingroup$ 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. $\endgroup$
    – Ant emyy Lee
    Commented Feb 11, 2010 at 2:47
  • $\begingroup$ 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. $\endgroup$
    – Ant emyy Lee
    Commented Feb 11, 2010 at 2:52

1 Answer 1

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Yes. The answer is obviously "yes" if $A$ is a subset of $g$, so it is sufficient to show for any subset $A\subset\kappa$ there is $A'\subset g$ such that $V[A']=V[A]$.

I assume that this is known, and it struck me at the start as obviously true, but I can't recall having seen it. Here is an outline of a proof.

Write the conditions (following Gitik) as $x=\langle x_i\mid i \in\omega\rangle$, with $x_i\in\kappa_i\cup U_i$. Write $A_n=A\cap \kappa_n$. There is $x\leq^* 1^P$ which forces that $A_n$ is decided by conditions $y$ with $y_i\in U_i$ for $i>n$. It follows in particular that $A_n\in V$.

Find $x'\leq^* x$ which decides, for each $n$, the sentence "there is $z\in \dot G$ such that $z_i$ decides the value of $\dot A_n$ and $z_i\in U_i$." Let $b_n$ be the (finite) set of $i$ for which this sentence is forced to be false. Then there is $x''\leq^* x'$ and 1--1 functions $h_n\colon \Pi_{i\in b_n} x_i''\to \kappa_n$ such that $x''$ forces that $A_n=h_n(g\upharpoonright b_n)$.

Set $A'=\bigcup_n b_n$. Then $V[A] = V[A']$.

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