Countable Σ¹₁ classes

1. Let $\{ X \} = \{ Z : \exists Y\, \varphi(Z, Y) \}$. Then $n \in X$ if and only if $\exists Z\, \exists Y\, (\varphi(Z, Y) \wedge n \in Z)$, a $\Sigma^1_1$ condition, and if and only if $\forall Z\, \forall Y\, (\varphi(Z, Y) \rightarrow n \in Z)$, a $\Pi^1_1$ condition, since $X$ is the only $Z$ with a witness $Y$. So $X$ is $\Delta^1_1$, hence hyperarithmetic by Kleene.
2. This is the second exercise of The hyperarithmetic sets are the Δ¹₁ sets: $X$ is hyperarithmetic if and only if there are $a$ and $e$ with $a \in \mathcal O$ and, for every $Y$ that is an H-sequence along $a$, $X = \Phi^{Y^{[a]}}_e$. By Lemma 2.2 of “The hyperarithmetic sets are the Δ¹₁ sets”, this expresses $X \leq_T H_a$, and the formula is $\Pi^1_1$.

Let $\mathcal B = \mathcal A \setminus \mathrm{HYP}$. It is a non-empty $\Sigma^1_1$ class, by Proposition 3.1, and the projection of the paths of a computable tree $S$ of pairs $u = (\sigma, \tau) \in 2^{<\nat} \times \nat^{<\nat}$ with $|\sigma| = |\tau|$. For such a pair, let $\mathcal B_u$ be the class of sets $X$ such that $(X, f)$ is a path of $S$ extending $u$ for some $f$; it is a $\Sigma^1_1$ class, with the finite object $u$ as a parameter, included in $\mathcal B$. Call $u$ good if $\mathcal B_u$ is non-empty. A good pair belongs to $S$, and the empty pair is good.

Splitting. Let $u$ be good. The class $\mathcal B_u$ is $\Sigma^1_1$ and contains only non-hyperarithmetic sets, so by Proposition 3.1 (1) it is not a singleton: it contains $X \neq X'$, with witnesses $f$ and $f'$. Let $k$ be such that $X(k) \neq X'(k)$, and $m > \max(k, |u|)$. Then $u_0 = (X \uh m, f \uh m)$ and $u_1 = (X' \uh m, f' \uh m)$ are good, extend $u$, and their first components are incompatible.

Construction. Define good pairs $u_s$ for $s \in 2^{<\nat}$ by $u_\emptyset = (\emptyset, \emptyset)$, and, given $u_s$, let $u_{s0}$ and $u_{s1}$ be the two pairs given by the splitting step. For $Z \in \cs$, the pairs $u_{Z \uh n}$ are increasing, and their lengths increase strictly, so they converge to a pair $(X_Z, f_Z)$. Every prefix of $(X_Z, f_Z)$ is a prefix of a good pair, hence belongs to the tree $S$; so $(X_Z, f_Z)$ is a path of $S$, and $X_Z \in \mathcal B \subseteq \mathcal A$.

The image is perfect. If $Z \neq Z'$, let $s$ be their longest common prefix; then $X_Z$ and $X_{Z'}$ extend the first components of $u_{s0}$ and $u_{s1}$, which are incompatible, so $X_Z \neq X_{Z'}$. The map $Z \mapsto X_Z$ is continuous, since $X_Z \uh n$ only depends on $Z \uh n$, because the first component of $u_{Z \uh n}$ has length at least $n$. Its image $\mathcal P$ is therefore compact, hence closed. It has no isolated point: for $Z \in \cs$ and $n \in \nat$, let $Z'$ differ from $Z$ only at position $n$; then $X_{Z'} \neq X_Z$, and both extend the first component of $u_{Z \uh n}$, of length at least $n$. So $\mathcal P$ is a non-empty perfect closed subclass of $\mathcal A$, in bijection with $\cs$.

1. This is the contrapositive of Theorem 3.2.
2. The class of graphs of the paths of $T$ is arithmetic, hence $\Sigma^1_1$, and in bijection with the set of paths; a path is hyperarithmetic if and only if its graph is. Item 1 gives the first statement. For the second, the class of graphs has a non-hyperarithmetic member, so it has the cardinality of the continuum by Theorem 3.2.