$\omega_2$-sequence of Suslin trees Is it possible to have an $\omega_2$-length sequence of ($\omega_1$-)Suslin trees such that if one builds the product of finitely many trees in that sequence, one ends up with a Suslin tree again?
The existence of such a sequence of length $\omega$ follows from $\diamondsuit$, as was shown by Jensen. By Shelah and independently Todorcevic, already a Cohen real gives rise to a Suslin tree, so it could be possible that a adding $\aleph_2$ many Cohen reals produces such a sequence.
 A: The answer is yes, and indeed, one can even have that any countable
number of the Suslin trees join to a Suslin tree.
To see this, simply force with countable support to add $\omega_2$
many Suslin trees. The forcing to add one Suslin tree has
conditions consisting of countable normal $\alpha$-tree, for
$\alpha<\omega_1$, and this tree will become an initial segment of
the desired generic tree. This forcing is countably closed and
isomorphic to $\text{Add}(\omega_1,1)$, adding a Cohen subset of
$\omega_1$. This forcing also adds a $\diamondsuit$-sequence, and
there is a tight connection between the argument from diamond that
there is a Suslin tree and the proof that this forcing adds a
Suslin tree. Namely, given any name for a maximal antichain, one
undertakes a bootstrapping argument to decide more and more of the
antichain, until one has a condition $t$ that decides $A\cap t$ and
such that $A\cap t$ is maximal in $t$. Then, one extends $t$ to
$\bar t$ with one more level in a way that seals that antichain in
any further extension of $\bar t$. So $\bar t$ forces that the
antichain is bounded and hence countable.
Consider now adding $\omega_2$ many such Suslin trees, with
countable support. An essentially similar sealing argument shows
that these trees are Suslin, and furthermore, that they are
mutually Suslin over each other. If you force with all but one of
these trees, with countable support, then the remaining tree will
still be Suslin. The reason is that if $\dot A$ is a name for
antichain in $T$, the unforced tree, where $\dot A$ is a name in
the product of the other trees, then it will be dense in the
product of the other tree-forcing conditions that forces that $\dot
A$ is bounded.
The argument is fundamentally similar to the methods used in my
paper


*

*Fuchs, Gunter; Hamkins, Joel David, Degrees of rigidity for Souslin trees, J. Symb. Log. 74, No. 2, 423-454 (2009). ZBL1179.03043. (blog post)


The point is that to add a generic Suslin tree $T$ and then force
with it, is the same as forcing with conditions $(t,b)$ where $t$
is a normal $(\alpha+1)$-tree and $b$ is an element on the top
level (determining the generic path). This forcing is very nice,
and can be used to show that a generic Suslin tree is Suslin off
the generic branch.
