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Let $M$ be a transitive model of ZFC (set- or class-sized) and let $\kappa\in M$ be a regular cardinal (in $V$). Let $G$ be $M$-generic for $\operatorname{Add}(\kappa,1)$. Now suppose that there is an $X\subseteq\kappa$ of size $\kappa$ which has no size $\kappa$ subsets in $M$ (for example, we could take $X=G$).

My question concerns which "patterns" can be realized in an $M$-generic on the coordinates in $X$.

Given a subset $Y\subseteq X$, is there an $M$-generic $G'$ for $\operatorname{Add}(\kappa,1)$ such that $G'\cap X=Y$? In other words, if we see $Y$ as a pattern of 0s and 1s, is there a generic $G'$ whose pattern on $X$ matches $Y$?

The requirement that $X$ has no large subsets in $M$ is clearly necessary: if $X'\subseteq X$ is in $M$ and has size $\kappa$, a density argument shows that $G'\cap X'\neq \emptyset$ for any generic $G'$, preventing us from realizing the all 0 pattern $Y=\emptyset$.

Certain patterns can be easy to achieve. For example, if we take $X=G$, then it is simple to realize the all 1 pattern $Y=G$: just take $G'=G$. Similarly, the all 0 pattern $Y=\emptyset$ is realized by $G'=\kappa\setminus G$. This last example also shows that it does not suffice, in general, to simply perform surgery on $G$ on the coordinates in $X$ and that global changes to $G$ might be necessary.

I am particularly interested in the special case when $M$ is the ultrapower by a measure on a measurable cardinal $\lambda$, we have $\kappa=\lambda^+$ and GCH holds. In this situation there is a variety of $M$-generics in $V$, which might make the problem easier.

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  • $\begingroup$ So, in a nutshell, you want to know which subsets of the generic factor as generic for a forcing whose remainder is in itself a Cohen forcing. Right? $\endgroup$
    – Asaf Karagila
    Commented Sep 26, 2016 at 16:17
  • $\begingroup$ @AsafKaragila Maybe? I'm not quite understanding your formulation. Can you explain? $\endgroup$ Commented Sep 28, 2016 at 1:55

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In some cases, it is possible to construct a set $X$ for which every pattern $Y\subseteq X$ is realized in the way you describe.

For example, consider the very natural case where $M$ is a countable transitive model of ZFC and $\kappa=\omega$. So we are talking about adding $M$-generic Cohen reals here. Enumerate the dense subsets $D_0,D_1,\ldots$ of $M$, and let us build a generic $G$ by diagonalization, while also constructing $X$ for which any $Y\subseteq X$ is realized. Construct a descending sequence of conditions $p_0\geq p_1\geq\dots$ as follows. Let $p_0$ be any element of $D_0$. Now, let the first element $i_0$ of $X$ be the first unspecified bit of $p_0$. Let $\bar p_1$ extend $p_0$ into $D_1$, and then flip the bit at $i_0$ in $\bar p_1$ and further extend to $p_1$ in $D_1$. So $p_1\in D_1$, extending $p_0$, and flipping the bit at $i_0$ keeps the condition in $D_1$. Continue similarly. At stage $n$, let $i_n$ be the first unspecified bit of $p_n$; first extend to $\bar p_{n+1}$ into $D_{n+1}$, and then flip the bits at all the $i_k$ for $k\leq n$ in any desired pattern, and extend into $D_{n+1}$ again, repeating for each pattern. At the end, let $p_{n+1}$ be the resulting extension of $p_n$, which has all its finite flips on the $i_k$'s in $D_{n+1}$. Let $G$ be the filter generated by the conditions $p_n$, which is $M$-generic since we have met every dense set, and let $X$ be the set of all $i_n$. By design, for any bit-flipping pattern on $X$, the construction exactly arranges that the resulting filter obtained by surgery will also be $M$-generic. Thus, every pattern $Y\subseteq X$ is realized by some $M$-generic surgical modification of $G$.

Update. Let me explain how a similar idea can work higher up, such as in the case of your ultrapower, with some additional assumptions.

Specifically, let's assume that $\kappa$ is a regular cardinal and $M$ is a transitive set or class model of ZFC, with $M^{<\kappa}\subset M$ and $|P(\kappa)^M|=\kappa$, which is true for your ultrapower example at the end of the question, if we have the GCH at your measurable cardinal. These assumptions are enough to construct $M$-generic Cohen sets $G\subset\kappa$ by the usual diagonalization procedure. You enumerate in $V$ the dense sets of $M$, and then construct a descending $\kappa$-sequence that gets inside them.

But let me now assume additionally that the $\Diamond_\kappa$ principle holds in $V$. In this case, I shall construct a set $X\subseteq\kappa$ such that every pattern $Y\subseteq X$ is realized in the way you request. Fix a $\Diamond_\kappa$-sequence $\langle A_\alpha\mid\alpha<\kappa\rangle$. Enumerate the dense sets of $M$ as $D_\alpha$, for $\alpha<\kappa$. At stage $\alpha$ of the construction, we have constructed a descending sequence of conditions $p_\beta$ for $\beta<\alpha$, and we have identified the first $\alpha$ many elements of $X$. Let $\bar p_\alpha$ be the union of the prior conditions, which is a condition in $M$ because $M^{<\kappa}\subset M$. We only do something special if the domain of $\bar p_\alpha$ is $\alpha$ itself, and if furthermore $A_\alpha$ is coding an ordinal $\beta<\kappa$ and a subset $Y$ of the first $\alpha$ members of $X$. In this case, we first extend $\bar p_\alpha$ into $D_\beta$, and then we further extend to a condition $p_\alpha$, so that after flipping the bits on $Y$, the condition remains in $D_\beta$. After this, let the $\alpha^{th}$ member of $X$ simply be the first bit unspecified by $p_\alpha$.

By design, the filter $G$ generated by the descending sequence of conditions $p_\alpha$ is an $M$-generic filter. If $Y\subseteq X$ is any set, then let $G_Y$ be the filter obtained by flipping the bits of $G$ on the coordinates of $Y$. This filter is also $M$-generic, since for any dense set $D_\beta$, there is a stationary set of stages $\alpha$ for which the domain of the conditions up to $\alpha$ is $\alpha$ itself (since that occurs on a club) and where $A_\alpha$ codes $\beta$ and the set $Y\cap\alpha$. In this case, the condition $p_\alpha$ had the property that flipping the bits on $Y$ remained in $D_\beta$, and so $G_Y$ meets $D_\beta$. Thus, any desired pattern can be arranged on the digits of $X$, by choosing $Y$ suitably.

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  • $\begingroup$ Joel, that is a great construction! I particularly like the argument from $\diamondsuit_\kappa$. However, your suggestion provides an $X$ and the generic $G$, while I was hoping that $X$ could be fixed in advance. Perhaps I shall edit the question to make it clear. $\endgroup$ Commented Sep 28, 2016 at 1:53
  • $\begingroup$ Oh, your question was clear, but I didn't have anything to say about that case. I think there can be some bad $X$, which don't realize all patterns, but I don't yet have an argument. $\endgroup$ Commented Sep 28, 2016 at 2:50
  • $\begingroup$ I am also wondering: what if you assume merely that $X$ itself is $M$-generic. In this case, can you realize every pattern $Y\subseteq X$? $\endgroup$ Commented Sep 28, 2016 at 2:51
  • $\begingroup$ Another natural case: assume $X$ decides every set in $M$ on a tail, so that the tail filter of $X$ is an $M$-ultrafilter. At first I thought this might realize all patterns, but I couldn't push an argument through. $\endgroup$ Commented Sep 28, 2016 at 3:06

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