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In ZFC, can we find more than continuum many non null sets of reals whose pairwise intersections are null?

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  • $\begingroup$ You have it already as a consistency assertion? $\endgroup$ Jun 12, 2015 at 11:55
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    $\begingroup$ Yes, e.g. under CH, there is a Sierpinski set so we can use an $\omega_2$ sized almost disjoint family on $\omega_1$ to construct such a family. $\endgroup$
    – Ashutosh
    Jun 12, 2015 at 11:58
  • $\begingroup$ Yes, I just came to a similar conclusion myself... $\endgroup$ Jun 12, 2015 at 11:59
  • $\begingroup$ Can't you do some sort of diagonalization on mutually disjoint Bernstein sets? $\endgroup$
    – Asaf Karagila
    Jun 12, 2015 at 12:05
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    $\begingroup$ I think this fails if continuum is real valued measurable. I will post details soon. $\endgroup$
    – Ashutosh
    Jun 14, 2015 at 11:09

1 Answer 1

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The answer is no. Observe that it is enough to show that, consistently, the density of the boolean algebra $\mathcal{P}(\mathbb{R}) / Null$ is continuum (which means that there are continuum many non null sets such that every non null set contains one of them). For this, it suffices to show that, consistently, ($\star$) holds:

($\star$): $2^{\omega} = \kappa = \kappa^{< \kappa}$ and every non null set of reals has a non null subset of size less than $\kappa$.

Claim 1: If $\kappa = 2^{\omega}$ is real valued measurable, then ($\star$) holds.

Proof: That $\kappa^{< \kappa} = \kappa$, is a result of Prikry (Theorem 22.2 in Jech's book). Let $\langle x_i : i < \kappa \rangle$ be a one-one enumeration of a non null set of reals $X$. Force with the null ideal of a measure on $\mathcal{P}(\kappa)$. Let $j: V \to M$ be the generic elementary embedding with critical point $\kappa$. Note that, in $M$, $X$ is a non null initial segment of $j(X)$ - this is because forcing with a measure algebra preserves old non null sets; so this also holds in $V$.

Claim 2: ($\star$) holds in the random real model.

Proof: (Communicated by Arnold Miller) Let $V \models GCH$, and $P$ add $\omega_2$ random reals $\langle r_i : i < \omega_2 \rangle$. Suppose $X = \langle x_i : i < \omega_2 \rangle$ is non null. Define $X_i = V[\langle r_j : j < i \rangle] \cap X$. Choose $\alpha < \omega_2$ of cofinality $\omega_1$ such that no null set coded in $V[\langle r_j : j < \alpha \rangle]$ covers $X_{\alpha}$. So $X_{\alpha}$ is non null in $V[\langle r_j : j < \alpha \rangle]$ and it remains so in $V[\langle r_i : i < \omega_2 \rangle]$.

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