3
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $x \in C$.

$\endgroup$
2
  • $\begingroup$ 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
    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. $\endgroup$ May 27, 2016 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.