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**=(x_{1},...,x_{n}). Φ is called *finitely satisfiable* if for every *finite* subset Ψ⊆Φ there exists **c**=(c_{1},...,c_{n}) such that ψ(**c**) holds for all ψ∈Ψ.

- 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 a_{1},...,a_{n}, then x=0.5*min{a_{1},...,a_{n}} satisfies __x<a___{1},...,__x<a___{n}. Thus Φ is finitely satisfiable.
- 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=n_{1}+...+n_{k} to satisfy the formulae indexed by these k n's.
- 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={φ
_{n}: n≥ 0}, where
φ_{n}(x) =
__∃t___{1}...∃t_{n}:
(t_{1}∈x & ... & t_{n}∈x) &
(t_{1}≠t_{2} & ... &
t_{1}≠t_{n} & ... & t_{n-1}≠t_{n}

Then Φ is finitely satisfiable, *even if our model includes only finite sets*, as long as it includes arbitrarily large finite sets.