Let $X$ be a Polish space and $\mathcal{F}(X)$ the set of closed subsets of $X$ endowed with the Effros Borel structure, generated by sets of the form $\{F\in \mathcal{F}(X):F\cap U\neq \emptyset\}$, where $U$ ranges over the open subsets of $X$.


Question: Is the set $\mathcal{C}(X)$ of all *clopen* subsets of $X$ a Borel subset of $\mathcal{F}(X)$? I am particularly interested in the case when $X=\mathbb{N}^\mathbb{N}$ (which is, notably, not locally compact).


One attempt to show this is to note that a closed set $F$ is clopen if and only if there is another closed set $G$ such that $F\cap G=\emptyset$ and $F\cup G=X$, but the intersection operation is not Borel on $\mathcal{F}(X)$ unless $X$ is locally compact, and even then, this appears at best a $\mathbf{\Sigma}^1\_1$, or analytic, characterization. Another thought is to use that the set $\mathcal{O}(X)$ of open subsets of $X$ can be endowed with a Borel structure via complementation... but I don't see how that helps (the intersection of Borel spaces need not be Borel if their structures don't cohere in some way).


**Edit:** If $X$ is totally disconnected, we can fix a sequence $(U\_n)$ of basic clopen subsets, an element of $\mathcal{F}(X)^\mathbb{N}$, and say that $F\in\mathcal{F}(X)$ is clopen if and only if for every $x\in F$, there is an $n$ such that $x\in U\_n$ and $U\_n\subseteq F$. The subset relation is Borel on $\mathcal{F}(X)$ (see section 12.C of Kechris, Classical Descriptive Set Theory), so this is a $\mathbf{\Pi}^1\_1$ description of being clopen. Together with what was written above, this shows that "clopen" is a Borel property in $\mathcal{F}(X)$, provided $X$ is totally disconnected and locally compact, but still leaves out the case $X=\mathbb{N}^\mathbb{N}$, in which I am most interested.



Is the set of clopen subsets Borel in the Effros Borel space?
Here is a negative answer for $\mathbb{N}^\mathbb{N}$.


Given a countably-branching tree $T$, we built a new countably-branching tree $T'$ in two steps. First, for any $\sigma \in T$ we place $(\sigma + 2)$ into $T'$, where $\sigma + 2$ is just obtained by adding $2$ to any number appearing in $\sigma$. In the second step, for any $\sigma$ already in $T'$, we add all words of the form $\sigma0\tau$ for $\tau \in \mathbb{N}^{<\omega}$ to $T'$.


So basically, we shift every branch in $T$ two steps to the right, and then add the full subtree below the $0$-child and nothing below the $1$-child. We can compute an enumeration of $T'$ from an enumeration of $T$, and we can compute the characteristic function of $T'$ from the characteristic function of $T$. Moreover, $T'$ definitely has no dead ends.


An enumeration of $T'$ thus constitutes a code of $[T'] \in \mathcal{F}(\mathbb{N}^\mathbb{N})$ [where I am not thinking about just the Borel-structure on this space, but the concrete representation of the space of overt (closed) subsets of Baire space].


If $T$ is ill-founded, then from any point $p \in [T]$ we obtain a point $(p+2) \in \delta([T'])$ [here $p+2$ is the result of pointwise adding $2$ to each number in $p$, and $\delta([T'])$ denotes the boundary of $[T']$], as $((p+2)\_{\leq n}0^\mathbb{N})\_{n \in \mathbb{N}}$ converges to $(p+2)$ inside $[T']$, and $((p+2)\_{\leq n}1^\mathbb{N})\_{n \in \mathbb{N}}$ converges to $(p+2)$ outside of $[T']$. This means that $[T']$ is not clopen.


On the other hand, if $T$ is well-founded, then we find that for any $p \in \mathbb{N}^\mathbb{N}$ there is some $n \in \mathbb{N}$ such that $p \in [T']?$ only depends on $p\_{\leq n}$, ie $[T']$ is clopen.


It follows that being clopen is $\Pi^1\_1$-hard for elements of $\mathcal{F}(\mathbb{N}^\mathbb{N})$. The very same construction also shows that being clopen is $\Pi^1\_1$-hard for elements of $\mathcal{A}(\mathbb{N}^\mathbb{N})$, too. [In the latter space, we code sets as paths through trees given via their characteristic function, with dead ends being allowed.]


