(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.

Log in or register to write something here or to contact authors.