Let f1 and f2 be two real-valued functions defined on sets X1,X2 respectively; then if A1,A2 are any non-empty subsets of X1,X2 respectively, we have (24)
(x1,x2)∈A1×A2sup(f1(x1)+f2(x2))=x1∈A1supf1(x1)+x2∈A2supf2(x2)
whenever both sides are defined; and if f1,f2 are ≥0 in A1,A2 respectively we have
(25)(x1,x2)∈A1×A2sup(f1(x1)f2(x2))=x1∈A1supf1(x1)x2∈A2supf2(x2),
whenever both sides are defined.
This is an immediate consequence of the preceding corollary and Proposition 9 of no. 4.
In particular, if A and B are two subsets of R such that the set A+B of sums x+y (x∈A,y∈B) is defined, we have
(26)sup(A+B)=supA+supB
if the right-hand side is defined. Again, if A and B are two subsets of [0,+∞], we have
(27)supAB=supA⋅supB
whenever both sides are defined.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.