Lemma 2 Let Bi and B2 be two sections of Rp. Then B = Bi U B2 is
also a section of Rp.
Proof Let B = Bi U B2 and assume that B1 and B2 are sections of Rp.
Let X, Y ∈ Rp be arbitrary. They can also be written as
X = (A ∩ B) U (A ∩ Bc)
and
Y = (У ∩ B) U (У ∩ Bc).
To show that B is a section, it is sufficient to show that (A ∩ B) U (У ∩ Bc) ∈
Rf . Rewrite X and Y as
X = (A ∩ (B1∖B2)) U (A ∩ (B2∖B1)) U (A ∩ (B1 ∩ B2)) U (A ∩ Bc)
and
Y = (У ∩ (B1∖B2)) U (У ∩ (B2∖B1)) U (У ∩ (B1 ∩ B2)) U (У ∩ Bc).
Since B1 is a section, (У ∩ (B1∖B2)) U (У ∩ (B1 ∩ B2)) and (A ∩ (B1∖B2)) U
(A ∩ (B1 ∩ B2)) belong to AC(Bi), and (У ∩ (B2∖B1)) U (У ∩ Bc) ∈ Cj1 ((У ∩
(B1∖B2)) ∩ (У ∩ (B1 ∩ B2))). Therefore,
(A ∩ (B1∖B2)) U (A ∩ (B1 ∩ B2)) U (У ∩ (B2∖B1)) U (У ∩ Bc) ∈ Rp.
By definition of the range complement of (У ∩ (B2∖B1)) U (X ∩ (B1 ∩ B2))
relative to B2,
(A ∩ (B1∖B2)) U (У ∩ Bc) ∈ ф((У ∩ (B2∖B1)) U (A ∩ (B1 ∩ B2))). (2)
Also, since X and Y belong to Rp and B2 is a section,
(A ∩ B2) U (У ∩ Bf ∈ Rp. (3)
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