Representations of positive projections 1 Introduction - Mathematics ...

Furthermore we de ne
B p = fpf : f 2 Bg : (14)
Then B p is a Dedekind complete regularly embedded f-subalgebra of A(p)
and it is easy to see that the Boolean algebra of components of p in B p is
given by
pCB = fpq : q 2CBg
which is a complete Boolean subalgebra of CA(p). If p 2 CB, then B p will
also be denoted by B(p).
For p 2CA we de ne
p =inffq 2CB : p qg , (15)
where the in mum is taken in CA. Since CB is a complete Boolean subalgebra
of CA, it follows that p 2CB. The element p is sometimes called the closure
of p with respect to CB ([10]). It is clear that p p.
Lemma 4.4 The mapping : B(p) ! B p, de ned by (f) = pf for all
f 2 B(p), is a surjective f-algebra isomorphism.
Proof. It is clear that is an f-algebra homomorphism. Now take
g 2 B p. Then g = pf for some f 2 B and hence (pf) = ppf = pf = g,
which shows that is surjective. It remains to proof that is injective. To
this end suppose that f 2 B(p) such that pf =0. From (15) it follows that
1 ; p =supfq 2CB : p ^ q =0g :
Since f = fp, i.e.,f (1 ; p) = 0, this implies that fq = 0 for all q 2CB \fpg d
and hence fs = 0 for all s 2 S(CB) \fpg d . Therefore, by Freudenthal's
spectral theorem we have fg =0for all g 2 B \fpg d . Since pf =0implies
that f 2 B \fpg d , it follows that f 2 =0,hence f =0,which shows that
is injective.
Remark 4.5 Note that the last argument in the above proof actually shows
that
B \fpg d(A) = B(p) d(B)
where d(A) and d(B) indicate that the disjoint complements are taken in A
and B respectively.
28