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By Yde Venema

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1. Let S = N, < be the bidirectional frame of the natural numbers with the standard ordering. That is, we interpret the diamonds ✸F and ✸P via the relations < and >, respectively. Furthermore, let A be the subalgebra of S+ based on the collection of finite and cofinite subsets of N. We claim that Var(A), the variety generated by A, has no complete members. Suppose for contradiction that C is a complete member of Var(A). Each natural number n is, inside S, the unique point satisfying the variable free formula ϕn := ✸nP ∧ ✷n+1 ⊥ (for m = n), and ϕn ✸F ϕn+1 hold in A, hence in P ⊥.

Of HBML), we obtain the following. Σ be a normal modal logic axiomatized by a collection Σ of Sahlqvist axioms. Then L is sound and complete with respect to the class of frames defined by the first-order correspondents of the formulas of Σ. 16 Although the Sahlqvist canonicity theorem takes care of most of the canonical formulas that one encounters in practice, it certainly does not cover the concept completely. For instance, G ORANKO & VAKARELOV [49] widen the class to that of so-called inductive formulas, see Chapter ??

Clearly p ≤ f σ (aq,0 , a1 ) for each q ∈ F ; since p is an atom this means p ≤ {f σ (aq,0 , a1 ) | q ∈ F } = f (a0 , a1 ), where in the last identity we use the additivity of f in its first coordinate. On the other hand, from x0 ≤ a0 and x1 ≤ a1 it follows that f σ (x0 , x1 ) ≤ f (a0 , a1 ) which gives the desired contradiction. ✷ In the proof above we already used the fact that complete additivity of f σ means that it is completely determined by its values on the atoms of Bσ . Now recall that (in the concrete representation of) Bσ , the atoms are nothing but the ultrafilters of B.

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