Page 43 - Textos de Matemática Vol. 41

5.4. The co-frame of sublocales 29 Use 1.3(H).)
Let ν be a nucleus and let ai = ν(bi) ∈ ν[L]. Then
ν(ai) ≤ ν(ai) = νν(bi) = ν(bi) = ai.
iiiii
If a is general and s = ν(b) ∈ ν[L] then a→s = ν(a→s) ∈ S by (5.3.3). Thus, ν[L] is a sublocale.
Let S be a sublocale. Then νS obviously satisfies (N1), (N2), (N3) and the inequality νS (a∧b) ≤ νS (a)∧νS (b). Now we have a∧b ≤ νS (a∧b), hence a ≤ b → νS(a∧b) and, by (S2), νS(a) ≤ b→νS(a∧b); further b ≤ νS(a)→νS(a∧b), and by (S2) again, νS(b) ≤ νS(a)→νS(a ∧ b) and finally νS(a) ∧ νS(b) ≤ νS(a ∧ b), by 1.3(H).
Furthermore νS[L] = S since νS(a) ∈ S by (S1), and for s ∈ S trivially νS(s) = s.
Finally,νSν(a)= {ν(b)|a≤ν(b), b∈L}=ν(a)sincea≤ν(a)andif a ≤ ν(b) then ν(a) ≤ νν(b) = ν(b).
5.4. The co-frame of sublocales. The set of all sublocales of L, ordered by inclusion, will be denoted by
Sl(L).
Obviously the least element in Sl(L) is O = {1} and the largest one is L.
Since all intersections of sublocales are sublocales, we have the infima in
Sl(L) given by Si. i∈J
The suprema in Sl(L) are given by the formula Si = {A | A ⊆ Si}.
i∈J i∈J
(The inclusion ⊇ is obvious. On the other hand, {A | A ⊆ Si} is a
i∈J sublocale: (S1) holds trivially, and, for any x ∈ L, x→ A = {x→a | a ∈ A}
by 1.3 (-distr), and each x→a is in i∈J Si.)
Proposition. Sl(L) is a co-frame (that is, the dual poset Sl(L)op is a frame).
Proof. Since the inclusion ( J Ai) ∨ B ⊆ J (Ai ∨ B) is trivial, we just have
to prove the reverse one. If x ∈ J(Ai ∨B) we have, for each i, ai ∈ Ai andb∈Bsuchthatx=a ∧b.Consequently,ifwesetb= b,wehave
iiii
x=( ai)∧b≤ai∧b≤ai∧bi =xsothatx=ai∧bforalli.By1.3.1(H6), x = b∧(b→ai) and by 1.3.1(H7) b→ai does not depend on i; denote its commonvaluebya.Thus,x=a∧bwithb∈Banda∈ JAi as,foreachi, a=b→ai ∈Ai.