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I am familiar with the result that $\Diamond$ implies that there exists a Suslin tree, but does it also imply that, if S, T are Suslin trees, then S x T is a Suslin tree? If not, perhaps there is a special class of Suslin trees where this nice result holds. Any input is welcomed.

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2 Answers 2

This is an interesting topic. The result is that you cannot really say much about $S\times T$ in general.

To be more precise, $\diamondsuit$ allows you to construct trees $S,T$ that are Suslin, but their product is special, or $S,T,U$ that are Suslin, $S\times T,S\times U,T\times U$ are Suslin, and $S\times T\times U$ is special. In fact, just about any pattern is possible.

An excellent reference for this and for techniques to use this freedom to "code information" is the paper

  • Uri Abraham, Saharon Shelah. "A $\Delta^2_2$ well-order of the reals and incompactness of $L(Q^{\rm MM})$." Ann. Pure Appl. Logic 59 (1993), no. 1, 1–32.
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The result is false in general since $S\times S$ is never Souslin. However, you can use $\diamondsuit$ to construct Souslin trees $T$ and $S$ such that its product is Souslin. You can find the construction in Todorcevic's paper in the Handbook of Set-Theoretic Topology.

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Hi, where can I find this construction? someone can paste the link here please? I couldn't find it –  luciana May 8 '12 at 23:21
I couldn't find the following: Todorcevic's paper in the Handbook of Set-Theoretic Topology. Can someone help me with this? I need to study the construction mentioned before about the product of two suslin trees using diamond. –  luciana May 9 '12 at 0:48

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