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Let $M$ be a transitive model of ZFC and $c\in {}^\omega2$ a Cohen real over $M$. Let $A$ be a meager Borel subset of $^\omega2$ in $M[c]$. I would like to prove that there exists a meager Borel set $B\subset{}^\omega2\times{}^\omega2$ with code in $M$ such that $A=B_c$.

I have seen this result stated in some texts, but I have not been able to complete a proof. Here, Andreas Blass gave me a hint to prove the analogous fact for null sets in random extensions, but I have failed to adapt it to the present case. Namely, I used there that the set $R(M)$ of random reals over $M$ has outer measure $0$ (although it is non-measurable). Here I would need that the set $C(M)$ of Cohen reals is comeager in $M[c]$, but I suspect this is false. At least, I cannot apply the 0-1 law to $C(M)$ if it does not have the Baire property.

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Here's a proof sketch: Let $p \Vdash \mathring{A} \subseteq 2^{\omega}$ codes a Borel set ($p$ is clopen).

Claim 1: There is a Borel set $B \subseteq 2^{\omega} \times 2^{\omega}$ coded in $V$ such that $p \Vdash B_{\mathring{c}} = \mathring{A}$ where $\mathring{c}$ is the name for the Cohen real.

Proof: First check this for open $A$ and then note that the family of sets $A$ for which the claim holds forms a sigma algebra.

Claim 2: Suppose moreover that $p \Vdash \mathring{A}$ is meager. Then we can take $B$ to be meager.

Proof: By Kuratowski-Ulam theorem, it is enough to show that $\{x \in p: B_x \text{ is not meager}\}$ is meager for the $B$ chosen above. Let $M$ be a countable transitive model of ZFC that contains a code for $B$. Let $C$ be the intersection of all open dense subsets of $p$ coded in $M$. Then $C$ is comeager and for every $x \in C$, $M[x] \models B_x$ is meager. By absoluteness, it follows that $V \models B_x$ is meager for every $x \in C$.

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  • $\begingroup$ Thank you!, but you are not choosing $B$ in any particular way, and in general, $B$ is not necessarily meager. Consider the trivial case in which $A=\emptyset$. You can take $p\subset B_0=\{x\in {}^\omega2 : x(0)=0\}$ and $B=B_1\times{}^\omega 2$. Then $p \Vdash B_{\mathring{c}} = \mathring{A}$, but $B$ is not meager. $\endgroup$
    – Carlos
    Commented Apr 23, 2016 at 14:05
  • $\begingroup$ I think the idea is that, given that $\{ x \in p : B_{x} \text{ is not meager}\}$ is meager, we can find a Borel meager set $C$ containing it, and then subtract $C \times 2^{\omega}$ from $B$. Then we can apply Kuratowski-Ulam to the Borel set $B \setminus (C \times 2^{\omega})$ to see that this set is what we want. $\endgroup$
    – Paul Larson
    Commented May 27, 2016 at 21:05

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