Suppose $M$ is a countable transitive model of ZFC and $(x, y, z)$ is Cohen generic point in $\mathbb{R}^3$ over $M$: This means that for every open dense set $U \subseteq \mathbb{R}^3$ in $M$, $(x, y, z) \in U$. I would like to know if $(x+y, x+z, y+z)$ is also Cohen generic point in $\mathbb{R}^3$ over $M$. I tried using product forcing lemma but couldn't show this. Neither could I get a counterexample. Thanks for any ideas!
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The function $\vec x\mapsto A\vec x$ defined by $(x,y,z)\mapsto(x+y,x+z,y+z)$ is an invertible linear operator (with inverse $(a,b,c)\mapsto\frac12 (a+b-c,a-b+c,-a+b+c)$) definable in the ground model $V$. Thus it maps dense open sets to and from dense open sets in the ground model, so for any dense open $U$ in $V$, $A(x,y,z)\in U\iff (x,y,z)\in A^{-1}U$, which is dense open and in $V$. Hence $A(x,y,z)$ is a generic point.
answered Oct 5, 2015 at 2:12
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$\begingroup$ This is almost right, but it's missing a crucial point: the function $\overline{x}\mapsto A\overline{x}$ you define exists in the ground model (and extends nicely to the generic extension, but that's clear): you need that it maps dense open sets in $V$ to and from dense open sets in $V$. For a silly example of why this matters: suppose $c$ is Cohen generic over $V$, and consider the linear map $f: x\mapsto {x\over c}$. Then $f$ is an invertible linear map, and $f(c)=1$, but clearly 1 isn't Cohen generic over $V$. What's going on is that $f$ exists and is nice in $V[c]$, but not $V$. $\endgroup$ Oct 5, 2015 at 2:20
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$\begingroup$ @NoahSchweber Yes, that is the missing piece. I'm not very good at forcing, so I tried to just apply the given definition. The fact that this is a "simple" map (which does not explicitly mention $(x,y,z)$) is clearly an important part of the proof, but I don't really have the background to quote the right theorems. $\endgroup$ Oct 5, 2015 at 2:34