Proof 2.

(a)

“”. Suppose is finite splitting. Let be a descending sequence of nonempty closed sets, where is a pruned subtree of .

(i) is infinite.

Proof. is obviously a tree. Suppose were finite. Let be an element of maximum length. Since , , for some , . Consider , since is finite splitting. For each , there exists such that . Let . Since , . But this is a contradiction, showing that is infinite.

(ii) is nonempty.

Proof. Since is infinite, and finite splitting, we must have . It is easy to see that . Hence, , so is compact.

I would prefer an argument based on trees. That is, suppose that K_i = [S_i]$ is a decreasing sequence of nonempty closed sets, where $S_i$ is a subtree of $T$ and show that $S = \Cap_i S_i$ is infinite and therefore [S] = \Cap [S_i]$ is nonempty.

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