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32 Chapter 6. Some special sublocales
(for c(a), (S1) is trivial and (S2) follows from 1.3.1(H4); for o(a), (S1) follows from 1.3 ( -distr) and (S2) from 1.3.1(H8): for a general y, y→(a→x) = a→ (y→x); as for the other formula, a→(a→x) = a→x by 1.3.1(H8) and the reverse inclusion is trivial).
The c(a) will be referred to as the closed sublocales and the o(a) as the open ones.
Realize that, however, while ↑a is a sublocale of L, ↓a is not. In fact we will see shortly that the o(a) is the (uniquely determined) complement of ↑a = c(a) and hence the open sublocales are necessarily defined as above.
On the other hand, recall 5.3.2 and 1.3.1(H7): the congruence created by the map
(x → a ∧ x) : L →↓a coincides with the congruence given by
(x → a → x) : L → o(a).
The former corresponds to the localic map ↓ a → L given by y → a → y which
can then be used to identify the sublocale embedding o(a) → L.
6.2.1. Nuclei and congruences of closed and open sublocales.
Proposition. We have
(1) νc(a)(x) = a ∨ x,
(2) νo(a)(x) = a→x,
(3) xEc(a)y iff x∨a=y∨a, (4) xEo(a)y iff x∧a=y∧a.
Proof. The first and the third formulas are obvious since νc(a)(x) = {s | x ≤ s,a ≤ s}.
Further,νo(a)(x)={a→y|x≤a→y, y∈L}.Wehavex≤a→xby 1.3.1(H4), and by 1.3.1(H8) if x ≤ a→y then
a→x ≤ a→(a→y) = a→y.
Finally, by 1.3.1(H7), a→x = a→y iff a ∧ x = a ∧ y.
6.2.2. Proposition. o(a) and c(a) are complements of each other in Sl(L).
Proof. If y = a→x ∈↑a we have a ≤ a→x, hence a ≤ x and y = a→x = 1 by 1.3.1(H3). Thus, o(a) ∩ c(a) = O. On the other hand, for any x ∈ L, x = (x ∨ a) ∧ (a→x) by 1.3.1(H9) so that x ∈ c(a) ∨ o(a).