Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 11 / 4 / a

Use a compatible complete metric on the Polish space $P$. For a tree on a Polish space $T\subseteq {\bigcup }_n{P}^n$, letbe its rank of a well-founded tree, with terminal rank zero. We bound this rank using a countable approximation tree, rather than assuming that the original tree has countably many nodes.

Choose a countable base of open balls. A node of the approximation tree $S$ at depth $n$ records, for every coordinate $i\le j\le n$, a basic ball $U_{j,i}$. Require its diameter to be less than $2^{-j}$, require $\overline{U_{j+1,i}}\subseteq U_{j,i}$, and requireInitial truncation removes the last row of balls. There are only countably many such finite records, so $S$ is countable.

An infinite branch in $S$ would give, by completeness and shrinking nested balls, a limit $p_i\in P$ for each coordinate. For fixed $j$, witnesses from later levels of $T$ have their first $j$ coordinates converging to $(p_1,\dots ,p_j)$. Initial-segment closure puts those witnesses in $T_j$, and closedness of $T_j$ puts the limit there too. Hence $(p_i)$ would be an infinite branch of $T$. Thus $S$ is well-founded.

Every tuple $t=(p_1,\dots ,p_n)\in T_n$ can be fitted inside such a record with each $p_i\in U_{n,i}$. Every extension $t^{⌢}p$ can be fitted in a child record: shrink the old balls around their actual points and choose a small ball around the new point. A well-founded induction therefore gives $r_T(t)\le r_S(s)$ whenever the record $s$ fits $t$. At the empty record this bounds the whole tree's rank by $r_S(\varnothing )$.

The rank of a countable well-founded tree is countable: at each node it is a supremum of countably many successors of countable ordinals, by well-founded recursion. Consequently a closed well-founded tree on a Polish space has countable height. The completeness and closedness assumptions are precisely what turn a branch of approximation records into an actual branch.