Erratum to "Hyperfinite and standard unifications for physical theories"

Definition 5.2. Suppose one has a nonempty finite set = {C1, . . . ,Cm} of general consequence operators, each defined on a nonempty language Li, 1≤ i≤m. Define the operatorΠCm as follows: for any X ⊂ L1×···×Lm, using the projection pri, 1≤ i≤m, define ΠCm(X)= C1(pr1(X))×···×Cm(prm(X)). Theorem 5.3. The operator ΠCm defined on the subsets of L1 ×···×Lm is a general consequence operator and if, at least, one member of is axiomless, then ΠCm is axiomless. If each member of is finitary and axiomless, then ΠCm is finitary. Proof. (a) Let X ⊂ L1 × ··· × Lm. Then for each i, 1 ≤ i ≤m, pri(X) ⊂ Ci(pri(X)) ⊂ Li. But, X ⊂ pr1(X)× ··· × prm(X) ⊂ C1(pr1(X))× ··· ×Cm(prm(X)) = ΠCm(X) ⊂ L1 × ···×Lm. Suppose that X = ∅. Then∅ =ΠCm(X)= C1(pr1(X))×···×Cm(prm(X))⊂ L1 × ··· × Lm. Hence, ∅ = pri(ΠCm(X)) = Ci(pri(X)), 1 ≤ i ≤ m, implies that Ci(pri (ΠCm(X)))= Ci(Ci(pri(X)))= Ci(pri(X)), 1≤ i≤m. Hence,ΠCm(ΠCm(X))=ΠCm(X). Let X =∅ and assume that no member of is axiomless. Then each pri(X) =∅. But, each Ci(pri(X)) = ∅ implies that ΠCm(X) = ∅. By the previous method, it follows, in this case, that ΠCm(ΠCm(X)) = ΠCm(X). Now suppose that there is some j such that Cj is axiomless. Hence, Cj(prj(X))=∅ implies thatΠCm(X)= C1(pr1(X))×···× Cm(prm(X)) = ∅, which implies that Cj(prj(ΠCm(X))) = ∅. Consequently, C1(pr1 (ΠCm(X)))× ··· ×Cm(prm(ΠCm(X))) = ∅. Thus, ΠCm(ΠCm(X)) = ∅ and axiom (1) holds. Also in the case where at least one member of is axiomless, then ΠCm is axiomless. (b) Let X ⊂ Y ⊂ L1 × ··· × Lm. For each i, 1 ≤ 1 ≤ m, pri(X) ⊂ pri(Y), whether pri(X) is the empty set or not. Hence, Ci(pri(X)) ⊂ Ci(pri(Y)). Therefore, ΠCm(X) =