Given a set $X$, by a *tree in $X$* I mean a set $T$ of finite sequences of elements of $X$ which is closed under initial segments. It is *pruned* of every element has a proper extension, and *finitely branching* if every element has at most finitely many immediate successors.

If $X$ is countable, then trees in $X$ are subsets of the countable set $X^{<\infty}$ of all finite sequences in $X$, and thus can be viewed as elements of the space $2^{X^{<\infty}}$, which is homeomorphic to the Cantor space $2^\mathbb{N}$ and thus Polish. Moreover, in this topology, the set of trees is a Borel subset of $2^{X^{<\infty}}$. Alternatively, if we consider only pruned trees, then these can be identified with their sets of infinite branches, which are exactly the closed subsets of the Polish space $X^\mathbb{N}$ (where $X$ is discrete), and thus inherits a standard Borel structure from the Effros Borel space of closed subsets of $X^{\mathbb{N}}$. The finitely branching pruned trees then correspond to the compact subsets of $X^{\mathbb{N}}$, and thus can be viewed as a compact Polish space under the Vietoris or Hausdorff topology on the space of compact subsets of $X^{\mathbb{N}}$.

**My question:** If we replace $X$ in the above by $\mathbb{R}$ (or any other uncountable Polish space), is there a natural standard Borel structure on the set of finitely branching pruned trees in $\mathbb{R}$?

My gut instinct is no, and probably one can use something like the fact that there is no nice standard Borel structure on the space of all countable subsets of $\mathbb{R}$, but I don't see how at the moment.

standardBorel, as in coming from a (separable) Polish topology. The issue is the finitely branching trees correspond to compact subsets of $X^{\mathbb{N}}$ where $X$ is discrete. $\endgroup$