(Robinson's Non-standard analysis)

While an infinite set of formulae may not be simultaneously satisfiable, it may still be possible to satisfy "many" of them.


Let Φ(x) be a set of formulae, each in the variables x=(x1,...,xn). Φ is called finitely satisfiable if for every finite subset Ψ⊆Φ there exists c=(c1,...,cn) such that ψ(c) holds for all ψ∈Ψ.


  1. In the language of real numbers with the standard model of the real numbers, the set of formulae "x<a", for every constant a∈R: Φ = {x:x<a: a∈R}. Φ is finitely satisfiable: if we are given the formulae indexed by a1,...,an, then x=0.5*min{a1,...,an} satisfies x<a1,...,x<an. Thus Φ is finitely satisfiable.
  2. In the language of natural numbers with the standard model of natural numbers, the set Φ = {x:x>n: n∈N}. Again, it is easy to see that Φ is finitely satisfiable -- e.g. take x=n1+...+nk to satisfy the formulae indexed by these k n's.
  3. In the language of set theory, let Φ be the set of formulae "x has at least n different elements", for all n. That is, Φ={φn: n≥ 0}, where
    φn(x) = ∃t1...∃tn: (t1∈x & ... & tn∈x) & (t1≠t2 & ... & t1≠tn & ... & tn-1≠tn
    Then Φ is finitely satisfiable, even if our model includes only finite sets, as long as it includes arbitrarily large finite sets.

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