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[A, B] = {X e J?: A < X < B> H = {(A, B) € Je : A ^ B> theorem 2.11 Let i/i g 0, then iJj is the supremum of a non-empty set of intervals. proof Take A, B g 1 such that A ^ B. Let <x(A, B) = Aty,(x): X g [A B]>. Consider the two operators e(A and e). defined as follows: (A,B) (A,B) C(A, B)(Z) ~ H I if Z = I a(A, B) if I >ZïA 0 if A | Z C(A, B)(Z) " H I if Z = 0 a(A, B) if O < Z ^ B 0 if Z | B e(A B) *S an eros^on anc* c(A B) *S an anti~erosi°n» then A(A, B) - e(A, B) A E(A, B) is the interval: (A, B) a(A, B) if A ^ Z ^ B 0 if Z | B or A I Z Let y = v"^(a g): € We must show that y = \fj. Take now (A, b) g H and Z g 1. If A ^ Z s b, then X(A b)(z) = a(A> B)* Hence> A B)(Z) s 0(Z). If A I Z or Z | b, then A^ gj(Z) = 0 ^ 0(Z). Thus A^ ^ \Jj(Z) for any (A, b) g K, and as A^ = i/»(Z), we have y(Z) = v*a(a b) : ^a' b^ € ^ = ^z^ Hence y = ifj and so ^ is a supremum of intervals. ■ There is also a dual result with an infimum of links. 2.5 Adjunctions We said above that dilation and erosion are dual concepts 22