Doesn't every nonempty tree on ${^k\omega} \times Y$ contain $\langle \emptyset, \emptyset, \dots, \emptyset \rangle$?

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This is a very small issue that I have when reading the chapter on "Sharps and $\Sigma^1_3$ Sets" from Kanamori's "The Higher Infinite".


To give some context from the book:

Suppose that $B \subseteq {^k(^\omega\omega)}$ is $\Sigma^1_2$. Taking $k = 1$ for simplicity, let $T$ be a tree on ${^2\omega} \times \omega$ such that for any $x \in {^\omega\omega}$, $$B(x) \leftrightarrow \exists^1y(T_{\langle x,y \rangle} \text{ is well-founded}).$$ $T$ can be defined from a relation recursive in a real $(13.1)$, and by slightly altering that relation if necessary it can be assumed that $\langle \emptyset, \emptyset, \emptyset \rangle \in T$.

This fact is stated again in different forms throughout the chapter.


Now I think by definition since any tree is downward-closed, if $T \not = \emptyset$ then we must have that $\langle \emptyset, \emptyset, \emptyset \rangle \in T$, no? And I don't get what changing the formula defining the tree has anything to do with this. I think I am missing some obvious and trivial point[because he mentions this in this chapter many times.] but I can't find it.