Terms and notation are defined here.

(Kechris exercise 4.11)

*(a) Let be a pruned tree on . It follows that is compact iff is finite splitting.*

* (b) In particular, if is compact, there is such that for all** , for every . Conclude that is not a countable union of** compact sets.*

*Proof.*

(a) “”. Suppose is not finite splitting. So there exists , of length , such that infinitely many satisfy . For each here, there exists such that . Pick sequence . Then, , for any . Hence, there is no convergent subsequence. Hence, is not compact.

“”. Suppose is finite splitting. Let be a sequence in . If , , let be the following statement: is an infinite

set. Then, it is clear that there exists such that holds, since there are only finitely many sequences of length in (since is finite splitting).

Now, assume inductively that there exists such that holds. Then, again since is finite splitting, there are only finitely many choices for . Hence, since is infinite, there must exist such that

holds.

Hence we have inductively defined a sequence such that for some . That is, contains a convergent subsequence, so is compact.

(b) Let be compact. Then is closed, and hence corresponds to a pruned tree , such that . So, for each , is finite (by the above). So define by . This satisfies the above property.

Now, suppose , where each is compact. Then for each we have such that implies for all . Now define by . Then, for any . Therefore, .

In part (a), you need to say that $y_{k+1}$ agrees with $y_k$ up to $k$, so that the sequence converges.

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.

Thank you Dr. Cenzer. Here is an alternate proof, along the lines you suggested.

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.