There is a (lightface) Σ<sup>0</sup><sub>1</sub> set A ⊆ ω such that for each p > 0 the Σ<sup>0</sup><sub>1</sub> set T<sup>p</sup> ⊆ ω<sup>p</sup> given by
> T<sup>p</sup>( j,x<sub>1</sub>,…,x<sub>p</sub> ) iff ∃t[ ⟨ j,⟨ x<sub>1</sub>,…,x<sub>p</sub>,t ⟩,1 ⟩ ∈ A ]
*parametrizes* the Σ<sup>0</sup><sub>1</sub> subsets of ω<sup>p</sup>, in the sense that X ⊆ ω<sup>p</sup> is Σ<sup>0</sup><sub>1</sub> iff for some j, X is the j-section
> { (x<sub>1</sub>,…,x<sub>p</sub>) : T<sup>p</sup>( j,x<sub>1</sub>,…,x<sub>p</sub> ) }
of T<sup>p</sup>. The set A is obtained by formalizing Kleene's notion of recursive derivation. (For details, see p. 127 of Moschovakis's *Descriptive Set Theory*, Second Edition. Any odd notation I use below is from that book; for instance, the asterisk will denote concatenation.)
We use A to define, for each pair p,n > 0, a set S<sup>p</sup><sub>n</sub> ⊆ ω<sup>p+1</sup> that parametrizes the Σ<sup>0</sup><sub>n</sub> subsets of ω<sup>p</sup>. It will be useful to write φ(α) as shorthand for the conjunction of this
> ∀j ∀y [ ∃i (α(i) = ⟨1,j,y⟩) ↔ (Seq(y) ∧ ∃t (⟨j,y*⟨t⟩,1 ∈ A)) ]
with this
> ∀j ∀y ∀m>0 [ ∃i (α(i) = ⟨m+1,j,y⟩) ↔ (Seq(y) ∧ ∃t ¬∃h (α(h) = ⟨m,j,y*⟨t⟩⟩)) ].
Here α ranges over <sup>ω</sup>ω and the Roman letters range over ω. Maintaining this convention, write ψ(α,n,j,y) for
> ∃m [ n = m+1 ∧ ∃i (α(i) = ⟨m+1,j,y⟩) ].
Notice that φ(α) ∧ ψ(α,n,j,y) defines an arithmetical subset of <sup>ω</sup>ω × ω<sup>3</sup>. Hence the set H ⊆ ω<sup>3</sup> given by
> H(n,j,k) iff ∃α (φ(α) ∧ ψ(α,n,j,y))
is Σ<sup>1</sup><sub>1</sub> since that pointclass is closed under projection along <sup>ω</sup>ω. Moreover, induction on n reveals that
> ∃α (φ(α) ∧ ψ(α,n,j,y))
is equivalent to
> ∀α (φ(α) → ψ(α,n,j,y))
so that H is in fact Δ<sup>1</sup><sub>1</sub>. Now for p,n > 0 define
> S<sup>p</sup><sub>n</sub> = { (j,x<sub>1</sub>,…,x<sub>p</sub>) : H(n,j,⟨x<sub>1</sub>,…,x<sub>p</sub>⟩) }.
By induction on n, for each p the set S<sup>p</sup><sub>n</sub> parametrizes the Σ<sup>0</sup><sub>n</sub> subsets of ω<sup>p</sup>. For the base, use the first conjunct of φ(α) to show that S<sup>p</sup><sub>1</sub> = T<sup>p</sup> for each p. For the inductive step, use the inductive hypothesis and the second conjunct of φ(α).
Finally, let Q ⊆ ω<sup>3</sup> be the Δ<sup>1</sup><sub>1</sub> set given by
> Q(n,j,k) iff H(n,j,⟨k⟩)
so that (n,j,k) ∈ Q iff (j,k) ∈ S<sup>1</sup><sub>n</sub>. If the foregoing is free of errors, this answers my original question.
The motivation for that question might have been obvious, but I'll put it down for the record.
> *The set Q witnesses that the arithmetical sets are not the "effective analog" of the Borel sets.*
A classical result of Suslin is that the (boldface) <b>Δ<sup>1</sup><sub>1</sub></b> sets coincide with the Borel sets. Since the arithmetical hierarchy resembles the Borel hierarchy, one might expect that the relationship between Δ<sup>1</sup><sub>1</sub> and arithmetical resembles that between <b>Δ<sup>1</sup><sub>1</sub></b> and Borel, enough perhaps that Δ<sup>1</sup><sub>1</sub> and arithmetical would coincide. It is well known that this expectation is false, and indeed the Δ<sup>1</sup><sub>1</sub> set Q witnesses this. For if Q were arithmetical, it would be Σ<sup>0</sup><sub>n</sub> for some n. Taking any Σ<sup>0</sup><sub>n+1</sub> set P ⊆ ω, there is some j such that for all k
> P(k) iff (j,k) ∈ S<sup>1</sup><sub>n+1</sub> iff (n+1,j,k) ∈ Q.
Since Q is Σ<sup>0</sup><sub>n</sub> by hypothesis, so is P. But then, since P was arbitrary, every Σ<sup>0</sup><sub>n+1</sub> subset of ω is in fact Σ<sup>0</sup><sub>n</sub>, contradicting the theorem that the arithmetical hierarchy is proper.