A proof of the compactness of first-order logic via ultrafilters, instead of the Henkin Construction.
Take a theory T which is finitely satisfiable. Hence for every finite Δ⊆T we know exists a structure MΔ⊨Δ.
Now let A be the collection of all Δ; have A={Δ⊆T}.
Now we are going to, in short, extend A to be a filter. For each Δ∈A let XΔ be all those supersets in T of Δ; expressly, let XΔ={Σ∈A:Σ⊇Δ}. And let our filter D be the upwards closure of all XΔ; that is,
D={Y⊆A:(∃Δ∈A)(Y⊇XΔ)}
I claim without proof—because I am fatigued—that D is a filter. Let U be some ultrafilter extension of D, and construct the ultraproduct
M=U∏Δ∈AMΔ
Now I claim that M⊨T and hence T is satisfiable, which is sufficient to complete the proof.
To show that M⊨T, we will show for each ϕ∈T that M⊨ϕ. So take ϕ∈T. Let
S={Δ∈A:MΔ⊨ϕ}
Now X{ϕ}⊆S, because for Σ∈X{ϕ} we know Σ⊇{ϕ} so MΣ⊨ϕ so Σ∈S.
Also X{ϕ}∈D, because {ϕ}∈A.
Thus S⊇X{ϕ}∈D⊆U, and so S∈U. Then by Los' theorem M⊨ϕ.