Lemma 3.1.6. Let L be a first-order language containing LCP L. Then there is a one-to-one correspondence between TarskianL-models and ultrafilters onLTL
CP Lthat preserve all universal quantifiers.
Proof. LetM= (D, J, α) be a Tarskian model, and letMφ be the set of all equivalence classes of true formulas in M. It is straightforward to check that Mφ is an ultrafilter ofLTL
CP L that
preserves all universal quantifiers. Conversely, if F is an ultrafilter on LTCP LL , then define
DF :=T erm(L)/∼F, αF : V ar(L)→ DF such that aF(x) = x
F for any x∈ V ar(L), and J F
such that for anyn-ary relation symbol P, JF(P) ={(aF1, ..., aFn) ; P(a1, ..., an)∈F}. Then it
is straightforward to check thatMF = (DF, JF, αF) is a TarskianL-model.
This fact is exactly what allowed Rasiowa-Sikorski to provide an algebraic proof of the com- pleteness of CPL with respect to Tarskian models, as we will see in the next section. However, before introducing the Rasiowa-Sikorski lemma for Boolean algebras, we conclude with a simple but far-reaching observation, that will allow us to use analogues of the Rasiowa-Sikorski lemma to obtain completeness results for logics other than CPL.
Lemma 3.1.7. Let C be a logic such that Cis sound and complete with respect to a variety of algebrasV such thatLTL
C ∈V, and such that the universal and existential quantifiers correspond
to infinitary meet and join in the sense of Lemma 3.1.3. Then for any classKof general models of C, C is sound and complete with respect to K if there is a term modelH ∈ K such that the valuationV associated to Kis injective.
Proof. Soundness is immediate, since if`Cφ, thenφ` =>`, which means that for any general
model (M, S(M), ..., V), V(φ`) = S(M). For completeness, it is enough to note that if H = (M, S(M), ..., V) is a term model in K andV is injective, then for any formula φ, ifφ` 6=>`, thenV(φ`)6=S(M), and hence if
0Cφ, then there exists a general modelHsuch that φis not
true onH, and henceφis not valid onK.
3.2
The Rasiowa-Sikorski Lemma for Boolean Algebras and
Completeness of CPL
In this section, we quickly recall the original proof of the Rasiowa-Sikorski Lemma1 using the
Stone representation Theorem. As it will soon become clear, all the generalizations of the lemma presented throughout this chapter are straightforward variations and refinements of this proof. The simplicity of the original proof, however, should not overshadow its originality not the important consequences that it has had for the study of first-order logic.
Lemma 3.2.1(Rasiowa-Sikorski Lemma). 2 Let B be a Boolean algebra, andQa countable set
of existing meets inB. Then for anya∈B, ifa6= 0, then there exists an ultrafilterU overB
such thata∈U andU preserves all meets inQ, i.e., for anyV
A∈Q, ifA⊆U, thenV
A∈U. Proof. Let (XB, τ) be the dual Stone space of B. Recall that for any a ∈ B,
|a| = {U ∈ XB ; a ∈ U}, and moreover that, by the ultrafilter theorem, for any a, b ∈ B,
a≤biff|a| ⊆ |b|. For anyV
A∈Q, letSA=Sa∈A−|a| ∪ |
V
A|. 1[55]
3.2. The Rasiowa-Sikorski Lemma for Boolean Algebras and Completeness of CPL
We first claim thatSA is open and dense in (XB, τ). It is immediate to see thatSA is open,
since|b|is always clopen for everyb∈B. For density, we show that
I(−SA) =I(
\
a∈A
|a| ∩ −|^A|)
is empty. To see this, recall first that
I(\
a∈A
|a| ∩ −|^A|) =I(\
a∈A
|a|)∩I(−|^A|).
Now assume that there is some basic open set|c|for somec∈B such that |c| ⊆T
a∈A|a|. This
means that |c| ⊆ |a|and hence that c≤afor any a∈A. But this clearly implies that c≤V
A
sincec is a lower bound ofA, from which it follows that|c| ⊆ |V
A|. Hence for any basic open set |c| ⊆T
a∈A|a|, we have that|c| ⊆ |
VA
|. This in turn implies by definition of the interior operator that I(\ a∈A |a|)⊆ |^A|. Hence I(−SA) =I( \ a∈A |a|)∩I(−|^A)⊆ |^A| ∩ −|^A|=∅.
ThereforeSA is dense in (XB, τ), since its complement is nowhere dense.
Moreover, recall that since (XB, τ) is a compact Hausdorff space, it satisfies the Baire Category
Theorem. Now let
SQ=
\
V A∈Q
SA.
By the previous result,SQis a countable intersection of open dense sets, and hence, by the Baire
Category Theorem, it is also dense in (XB, τ).
The conclusion of the lemma follows at once. For anya∈XB, ifa6= 0, then|a| 6=|0|=∅. But
then, since|a| is always open andSQ is dense,|a| ∩SQ is nonempty. Clearly, anyU ∈ |a| ∩SQ
is an ultrafilter that preserves all meets inQand containsa.
The main consequence of the previous lemma is the following completeness proof:
Theorem 3.2.2(Rasiowa and Sikorski). CPL is complete with respect to the class of all Tarskian models.
Proof. Let L be a countable first-order language. By Lemma 3.1.6, any ultrafilter on LTCP LL
that preserves all meets in
Q∀={(∀xφ(x))`; φ∈L}
gives rise to a Tarskian term model of CP L. Now for any φ∈L, if 0CP Lφ, then (¬φ)` 6=⊥`.
But then, by the Rasiowa-Sikorski lemma, since Q∀ is countable, there exists an ultrafilter U overLTL
CP L that preserves all meets inQ∀ and does not contain φ. Hence the Tarskian model MU corresponding to U is such that φis not true onMU, which means thatφ is not valid on
the class of all Tarskian model. By contraposition, it follows thatCP Lis complete with respect to the class of all Tarskian models.