# Silver forcing and Cichoń's diagram

Recall that the Silver forcing $\mathbb{P}$ is defined as the set of all partial functions $p\in 2^{\le\omega}$ such that $\omega\setminus dom(p)$ is infinite. As usual, $p\le_\mathbb{P}q$ if $p$ extends $q$. The Silver model is the model $V$ obtained by the countable support iteration of length $\omega_2$ of the Silver forcing over a model of ZFC+CH.

I'm interested in values of the cardinal characteristics of the continuum in the van Douwen and Cichoń diagrams in $V$. Halbeisen in his brilliant book "Combintorial Set Theory" proves that:

$V\models \omega_1=\mathfrak{d}<\mathfrak{r}=\mathfrak{c}=\omega_2$.

It follows immediately that $V\models\mathfrak{u}=\mathfrak{i}=\omega_2$ and that all characteristics of the van Douwen diagram below $\mathfrak{d}$ are equal to $\omega_1$ in $V$. The only unknown in the diagram is the almost-disjointness number $\mathfrak{a}$.

Question 1: What is the value of $\mathfrak{a}$ in $V$?

Let $\mathcal{M}$ and $\mathcal{N}$ denote the ideal of meager subsets and the ideal of Lebesgue null subsets of $\mathbb{R}$, respectively. It can be shown that the ground model reals are non-meager in $V$, so $non(\mathcal{M})=\omega_1$ in $V$, and hence the left-hand half of the Cichoń diagram is also equal to $\omega_1$. Since $\mathfrak{d}=\omega_1$ in $V$, $cov(\mathcal{M})=\omega_1$ in $V$ as well. The rest of the Cichoń diagram is unfortunately unknown to me.

Question 2: What are the values of $cof(\mathcal{M})$, $non(\mathcal{N})$ and $cof(\mathcal{N})$ in $V$?

• Doesn't the Silver forcing have the Sacks property? I.e., every new function in $\omega^\omega$ is contained in an old slalom. That should imply the every new null set is contained in an old null set. – Goldstern Dec 27 '17 at 14:13
• It has! (It's $\omega^\omega$-bounding and has the Laver property -- proofs can be found in Halbeisen's book) And in Geschke and Quickert's survey "On Sacks forcing and the Sacks property" one can found a proof that the Sacks property implies that $cof(\mathcal{N})$ has the value of the ground model $2^\omega$. Thank you, Martin! (However, that's sad news for me...) – Damian Sobota Dec 27 '17 at 15:11
• The arguments using Sacks forcing in the following should be able to be adapted to give $\mathfrak{a}=\aleph_1$ in the Silver model. gaspacho.matmor.unam.mx/~michael/reprints_files/… – Iian Smythe Dec 27 '17 at 17:26
• Alternatively, since $\mathfrak{b}=\aleph_1$ in the Silver model, it satisfies $\diamondsuit(\mathfrak{b})$ which implies $\mathfrak{a}=\aleph_1$, see: math.cornell.edu/~justin/Ftp/diamonds.pdf – Iian Smythe Dec 27 '17 at 17:32
• I converted my comment into an answer. Halbeisen's book is indeed very nice, but I added a few more references (namely, those that I grew up on). – Goldstern Dec 29 '17 at 12:54

Answer to question 2: $cof(\mathcal N)=\aleph_1$, so all cardinals in Cichoń's diagram stay small (and their smallness is witnessed by the set of reals from the ground model).
Proof sketch: For $q\le p$ in Silver forcing, write $q\le_n p$ if $\omega\setminus dom(p)$ and $\omega\setminus dom(q)$ agree on their first $n$ elements. This is an axiom A structure with the following "pure extension" property: Whenever $\dot \alpha$ is a name of an ordinal, $p$ a condition, $n\in \omega$, then there is a set $A$ of size $2^n$ and a condition $q\le_n p$ such that $q\Vdash \dot \alpha\in A$.
This shows that for every new function (in the single Silver extension), $f:\omega\to\omega$ there is an old slalom $(S_n:n\in\omega)$, $|S_n|=2^n$, $\forall n: f(n)\in S_n$, in other words: the single Silver extension has the Sacks property.
The Sacks property is preserved in countable support iterations of proper forcings. (Shelah, Proper and Improper Forcing, chapter XVIII.) But for an iteration of Silver forcing, or any sufficiently nice Axiom A forcing, this can alternatively be proved "directly", with the usual $\le_{F,n}$-fusion argument. (E.g. in Baumgartner's classical paper "Iterated Forcing", if I remember correctly.)
Hence also the model after adding $\omega_2$ Silver reals has the Sacks property over the ground model.