I'm trying to prove exercise 36 From Kunen "Set Theory", Chapter II:
If $T$ and $T'$ are $\kappa$-trees, define the product $T\times T'$ to be the $\kappa$-tree whose $\alpha$-th level is $\operatorname{Lev}_\alpha (T) \times \operatorname{Lev}_{\alpha}(T')$ with order defined by $(x,x') < (y,y')$ iff $x<y$ and $x'<y'.$
Show that if $T$ is a $\kappa$-Suslin tree, then $T\times T$ is not a $\kappa$-Suslin tree.
In the process I thought about this argument: Let C be a maximum maximal anti-chain in $\kappa$-suslin tree T, and assume $T_x=\left\{z\in T:z\le x\vee x\le z\right\}$. If $\left|\bigcup_{x\in C} T_x\right|\neq\kappa$, there is $b\in\bigcap_{x\in C} T_x^c$, such that b is incompatible to any $x\in C$. Therefore, because $\kappa$ is regular $\left|C\right|=\kappa$ and there is no $\kappa$-suslin tree. What is wrong here?
My approach for the exercise is to show that there is a level or chain with no element from C from regularity, and find another element of T from the same level such that they are incompatible with any other elements of C and therefore contradict the fact that the maximum anti chain is from cardinality less than $\kappa$. can anyone give me a hint/way to solve this question?
As Andreas mentioned in the comments, you seem to be assuming $|T_x|<\kappa,$ but that's not true... in fact if the tree is well-pruned $|T_x| = \kappa$ for all $x\in T.$ (Just because there is no branch of height $\kappa$ doesn't mean a given node doesn't have descendents on every level through $\kappa$.)
As to how to do it, first off, the exercise is stated incorrectly relative to Kunen's definitions: we must assume $\kappa$ is a regular cardinal or there are simple counterexamples. Actually, we can show the following stronger statement:
To show this, first pass to a well-pruned subtree (an antichain in the subtree will be an antichain in the original tree). Then recursively choose distinct $a_\alpha,$ $b_\alpha,$ $c_\alpha\in T$ for $\alpha<\kappa$ such that $a_\alpha, c_\alpha >_T b_\alpha,$ $\operatorname{ht}_T(a_\alpha) = \operatorname{ht}_T(c_\alpha)$ and $\operatorname{ht}_T(b_\alpha) > \operatorname{ht}_T(a_\xi), \operatorname{ht}_T(c_\xi)$ for all $\xi < \alpha.$ Then, observe $\{(a_\alpha, c_\alpha): \alpha < \kappa\}$ is an antichain of size $\kappa$ in $T\times T.$