3 deleted 5 characters in body

For each n, there is a (lightface) Σ0n set Sn ⊆ ω2 that's universal for the Σ0n subsets of ω. Since {n} × Sn is Σ0n, there is a union R of arithmetical sets such that (n, j, k) ∈ R iff (j, k) ∈ Sn. Clearly R is not itself arithmetical, and offhand I don't see why it should be even Δ11.

If we define the sets Sn with care, is there a Δ11 set Q ⊆ ω3 such that (n, j, k) ∈ Q iff (j, k) ∈ Sn?

2 Corrected blunders about Borel sets.

For each n, there is a (lightface) Σ0n set Sn ⊆ ω2 that's universal for the Σ0n subsets of ω. Since {n} × Sn is Σ0n and the Borel sets are closed under countable union, there is a Borel set union R ω3of arithmetical sets such that (n, j, k) ∈ R iff (j, k) ∈ Sn. Clearly R is not itself arithmetical, and offhand I don't see why it should be even Δ11.

If we define the sets Sn with care, is there a Δ11 set Q ⊆ ω3 such that (n, j, k) ∈ Q iff (j, k) ∈ Sn?

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# Code universal arithmetical sets by a hyperarithmetical set?

For each n, there is a (lightface) Σ0n set Sn ⊆ ω2 that's universal for the Σ0n subsets of ω. Since {n} × Sn is Σ0n and the Borel sets are closed under countable union, there is a Borel set R ⊆ ω3 such that (n, j, k) ∈ R iff (j, k) ∈ Sn.

If we define the sets Sn with care, is there a Δ11 set Q ⊆ ω3 such that (n, j, k) ∈ Q iff (j, k) ∈ Sn?