truth of a sentence to the linearly ordered set Let Φ - sentence of signature σ = <≤> such that for any infinite linearly ordered set A satisfies A ⊨ F. Prove that there exists n ∈ N such that for every linearly ordered set B power greater than n, we have B ⊨ Φ
 A: HINT: Assume by contradiction, and use either ultraproducts or compactness to construct an infinite linear order which does not satisfy $\Phi$.
A: We assume that you are working in the predicate calculus with equality. Let $T_0$ be the theory of linear order.
Suppose there are arbitrarily large finite linearly ordered sets that satisfy $\lnot F$. Let $\varphi_n$ be the usual sentence that says there are at least $n$ elements.
Let $T$ be the theory whose axioms are $T_0$, together with $\lnot F$, together with the collection of all $\varphi_n$. By our assumption, every finite subset of $T$ has a model. Thus by Compactness $T$ has a model. This model cannot be finite because of the presence of all the $\varphi_n$. In this model, $\lnot F$ is false, contradicting the fact that $F$ is true in every infinite linearly ordered set. 
Remark: Note that we nowhere used the fact that the theory $T_0$ is the theory of linear order. 
A: Hint: use Ehrenfeucht-Fraisse games to show that for every $n$ there is an infinite linearly ordered set and a finite linearly ordered set satisfying the same sentences with $n$ quantifiers (in fact, you can just pick a single infinite linear order and just adjust the finite one according to $n$).
