Evaluating elementary Borel codes

Let $T_1=\{\mathrm{\varnothing }\}$ with root label leaf$(n)$. It is valid since it has no child and its empty relation is well-founded. F2 gives $E_1(\mathrm{\varnothing })=U_n$. For $T_2=\{\mathrm{\varnothing },(0)\}$ label the root complement and $(0)$ leaf$(n)$. Then $E_2((0))=U_n$ and $E_2(\mathrm{\varnothing })=X\setminus U_n$, which is closed because $U_n$ is open.

On $T_3=\{\mathrm{\varnothing }\}$ put a union label; then $E_3(\mathrm{\varnothing })=\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$. On $T_4=\{\mathrm{\varnothing },(0)\}$ put a complement at the root and a union at $(0)$. Then $E_4((0))=\mathrm{\varnothing }$ and $E_4(\mathrm{\varnothing })=X\setminus \mathrm{\varnothing }=X$.

In each two-node tree the sole child is terminal; any subset containing it has it as a minimal element, and a remaining nonempty subset is the singleton root. Thus the child relation is well-founded, and the complement nodes have exactly child zero as required. All four codes are valid even when $X=\mathrm{\varnothing }$ or $U_n$ is empty or all of $X$. The empty underlying tree is not one of these codes, because F1 requires a root. QED.