By definition, a tree is a set of finite sequences on some alphabet $A$ (equipped with the discrete topology) which is a lower set w.r.t. the initial segment relation. A tree is said to be pruned iff every finite sequence in the tree has a proper extension in the tree. Finally, the body of a tree $T$ is $$[T]=\{x\in A^{\mathbb{N}}\colon \forall n(x|n\in T)\}.$$
Kechris in his "Classical Descriptive Set Theory" pp. $7$, Prop $(2.4)$ states that:
The map that associates the body of a tree to each tree is a bijection between pruned trees and closed subsets of $A^{\mathbb{N}}$ (equipped with the product topology).
I have proved that it is a bijection, but I want to show that the body of a tree is necessarily closed. I have tried to prove that the complement is open, but I don't succeed.
Any suggestion will be appreciated.
Edit: I found the following answer (I hope it is correct) some time ago, but I would like to see other methods to prove its closedness.
Let $N_{x|n}=\{y\in A^\Bbb{N}\mid x|n\,\text{jnitial segment of}\,y\}$ be an open nhbd of $x$, where $n$ is the least whole number for which $x|n\notin T$ (here $x|n=(x_1,\dots,x_n)\in A^n$ whenever $x=(x_1,\dots,x_n,x_{n+1},\dots)\in A^\Bbb{N}$). Now, $N_{x|n} \subseteq [T]^c$ and hence $[T]^c=\bigcup_{x\in [T]^c} N_{x|n}$.
Your proof in itself is fine: the complement is open, using the basic open sets from initial segments.
But maybe easier: note that for any $n$: $\pi_n: A^{\mathbb{N}} \to A^n$ is continuous (as all projections are in the product topology) and $A^n$ is discrete as a finite product of discrete spaces, which makes $T_n:= T \cap A^n$ (the $n$-th level of $T$) closed in $A^n$, trivially.
Finally note that $$[T]=\bigcap_{n \in \mathbb{N}} \pi_n^{-1}[T_n]$$
is an intersection of closed sets and thus closed in $A^\mathbb{N}$.