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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? |
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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?
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