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I just read a quite sketchy proof, where some details were skipped and it appears to me that the proof uses some of the following things. However a (probably very easy) question came up:

Assume that we force with a Boolean algebra of the form $P(A) / I$, where $A$ is an arbitrary set and $I$ is an ideal over $A$; i.e. our forcing conditions are $I$-postive subsets of $A$ and $p \le q$ iff $p \supset q$.

Moreover assume that $P(A) /I$ forces that $\dot{f}$ is a function in $V$ from $A$ to $V$ (where $V$ denotes our ground model), then it follows that there exists a maximal antichain $(S_{\alpha} : \alpha < \lambda)$ in $P(A)/I$ and corresponding functions (lying in $V$ ) $g_{\alpha}$:$A$ $\to V$ such that $S_{\alpha} \Vdash \dot{f} = g_{\alpha}$.

Now if the antichain can be written in a way such that the representatives of the $S_{\alpha}'s$ are even pairwise disjoint then the partial functions $g_{\alpha}\upharpoonright S_{\alpha}$ can be glued together to a new function $g$ and now it should follow that $P(A) / I$ forces that $\dot{f} = g$.

My question now is: How does one see that the $P(A) /I $ forces $\dot{f} =g$ ?

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    $\begingroup$ I don't think your gluing fact is true in the generality you've given. For example, do random forcing and take $\dot{f}$ to name the function which is constantly whatever value the adjoined random real takes at 0. Gluing will net you a $g$ which isn't a constant function. $\endgroup$
    – Justin Palumbo
    Commented Apr 6, 2011 at 15:57
  • $\begingroup$ If you're right, Justin then this would explain why I had a hard time proving it. The above 'gluing fact' seems to appear in a lemma in the lecture of my professor, and no one of the sophisticated audience said anything against it. However Jech's book (currently my only source of knowledge at hand) has to say very little about Random Forcing, so I would be glad if you could explain your comment a little bit. $\endgroup$
    – Stefan Hoffelner
    Commented Apr 6, 2011 at 16:49
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    $\begingroup$ Take $A$ to be $2^\omega$, and let $I$ be the ideal of all measure zero sets. This is random forcing. Let $X$ be the collection of all members of $2^\omega$ whose first digit is 0, and $Y$ be all those whose first digit is 1. Exactly one of these lie in any generic filter $G$. Let $\dot{f}$ name the function which is constantly 0 if $G$ contains $X$, and constantly 1 if $G$ contains $Y$. Then $X$ and $Y$ together form a maximal antichain, but if you do the gluing process you describe, you get a function which is 0 on $X$ and 1 on $Y$, and so cannot equal $\dot{f}$, whatever it ends up being. $\endgroup$
    – Justin Palumbo
    Commented Apr 6, 2011 at 17:06
  • $\begingroup$ (by the way this argument has almost nothing to do with random forcing, and will work for essentially any A and I) $\endgroup$
    – Justin Palumbo
    Commented Apr 6, 2011 at 17:17

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The generic subset $G\subseteq P(A)/I$ adjoined by your forcing can be viewed as an ultrafilter on $A$ (disjoint from the ideal $I$). Your glued-together $g$ will (be forced to) be equal to $f$ on a set in this ultrafilter; that is, $g$ and $f$ will represent the same element of the generic ultrapower of $V$. But, as Justin Palumbo noted, they won't generally be equal everywhere.

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  • $\begingroup$ Now, with your help I see that it was just a notational issue: The brackets around $f$ and $g$ (denoting the elements in the generic ultrapower) were missing. Thank you very much. $\endgroup$
    – Stefan Hoffelner
    Commented Apr 6, 2011 at 18:50

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