Page 95 - Textos de Matemática Vol. 41

12.3. The relations 1E and 2E 81
and let us prove that j0(E) ∈ A whenever E ∈ A. Consider (x, y) ∈ j0(E) · B andz̸=0suchthat(x,z)∈j0(E)and(z,y)∈B.If(x,z)=(x, S)forsomeS w i t h { x } × S ⊆ E , t h e r e i s a n o n - z e r o s ∈ S s u c h t h a t ( x , s ) ∈ E a n d ( s , y ) ∈ B and,therefore,(x,y)∈E·B⊆j(A·B).Ontheotherhand,if(x,z)=( S,z) for some S with S × {z} ⊆ E, (s, y) ∈ E · B for every s ∈ S and, therefore, (x, y) ∈ j0(E · B) ⊆ j(A · B).
Moreover, for any non-void F ⊆ A, F ∈F F ∈ A, since ( F)·B⊆ (F·B).
F∈F F∈F
Therefore S = E∈A E belongs to A, i.e., A has a largest element S. But j0(S) ∈ A so S = j0(S), that is, S is saturated. Hence j(A) = S ∈ A and, consequently, j(A) · B ⊆ j(A · B). By symmetry, A · j(B) ⊆ j(A · B).
In conclusion, we have
j(A) · j(B) ⊆ j(A · j(B)) ⊆ j2(A · B) = j(A · B).
The results above mean that
(L ⊕ L, ◦) is a quantic quotient [46] of the quantale (D(L × L), ·).
Indeed: j0 is a quantic prenucleus on the quantale (D(L × L), ·), that is,
(1) A ⊆ j0(A),
(2) if A ⊆ B then j0(A) ≤ j0(B),
(3) j0(A)·B≤j0(A·B)andA·j0(B)≤j0(A·B),forallA,B∈D(L×L).
Then, the associated closure operator j is, by Lemma 12.2.5, a quantic nucleus on (D(L × L), ·), that is, j(A) · j(B) ⊆ j(A · B). This implies, by general results on quantales, that Fix(j) = L⊕L is a quantale with joins jEi = j(Ei) and multiplicationE·j F =j(E·F).ButbyLemma12.2.5E·j F =j(E)◦j(F)= E ◦ F and, therefore, (L ⊕ L, ◦) is a quantale.
In particular, this asserts that the multiplication ◦ of (12.12.1) is associa- tive.
12.3. The relations 1E and 2E . For a system E of entourages on L define the relations 1E and 2E by, respectively,
and
b 1E a ≡ ∃ E ∈ E , E ◦ ( b ⊕ b ) ⊆ a ⊕ a b 2E a ≡ ∃ E ∈ E , ( b ⊕ b ) ◦ E ⊆ a ⊕ a .