I'm reading a portion of Donald A. Martin's book manuscript. On page 56, it is mentioned,
I'm unsure what $\lceil{T}\rceil$ is. It is probably defined in Chapter $1$ which hasn't yet been posted. I guess it's the union of the set of all infinite plays and the set of terminal positions. I don't know what topology is defined on $\lceil{T}\rceil$ either.
If $\lceil{T}\rceil$ is the union of the set of all infinite plays and the set of terminal positions, then the notation $G(A \cup \mathcal{T}_{\text{II}}\setminus \mathcal{T}_{\text{I}}; T)$ on page 56, is weird to me, since $G(A \cup \mathcal{T}_{\text{II}}; T)$ would suffice($A$ and $\mathcal{T}_{\text{II}}$ are both disjoint from $\mathcal{T}_{\text{I}}$).
Question: Why is $[T]$ closed in $\lceil{T}\rceil$?
The material at the top of page $57$ indicates that you’re right in thinking that $\lceil T\rceil$ must be the union of $[T]$ and the set of terminal positions: $f$ sends the infinite play $p^{\frown}000\ldots$ in $T'$ to the terminal position $p$ in $T$. Since $f$ is stated to be a homeomorphism, I think it extremely likely that the topology on $\lceil T\rceil$ is the one for which the sets $B_p=\{x\in\lceil T\rceil:x\supseteq p\}$ for $p\in T$ form a base. Note that if $p$ is a terminal position, then $B_p=\{p\}$, so $p$ is an isolated point of $\lceil T\rceil$, and it follows immediately that $[T]$ is closed in $\lceil T\rceil$.
I don’t know why Martin chose to write $G((A\cup\mathcal{T}_{\text{II}})\setminus\mathcal{T}_{\text{I}};T)$ instead of $G(A\cup\mathcal{T}_{\text{II}};T)$; it may well be that $A\subseteq[T]$ is a typo for $A\subseteq\lceil T\rceil$, in which case the definition makes sense as it stands.