Bourbaki

Corollary 2

Tag 0BRT, permanent. Cite it as [Bourbaki, Tag 0BRT].

General Topology, chapter IV, § 5. Real-valued functions

Let f1f_1 and f2f_2 be two real-valued functions defined on sets X1,X2X_1, X_2 respectively; then if A1,A2A_1, A_2 are any non-empty subsets of X1,X2X_1, X_2 respectively, we have (24)

sup⁡(x1,x2)∈A1×A2(f1(x1)+f2(x2))=sup⁡x1∈A1f1(x1)+sup⁡x2∈A2f2(x2) \sup_{(x_1, x_2) \in A_1 \times A_2} (f_1(x_1) + f_2(x_2)) = \sup_{x_1 \in A_1} f_1(x_1) + \sup_{x_2 \in A_2} f_2(x_2)

whenever both sides are defined; and if f1,f2f_1, f_2 are ≥0\geq 0 in A1,A2A_1, A_2 respectively we have

(25)sup⁡(x1,x2)∈A1×A2(f1(x1)f2(x2))=sup⁡x1∈A1f1(x1)sup⁡x2∈A2f2(x2), (25) \quad \sup_{(x_1, x_2) \in A_1 \times A_2} (f_1(x_1)f_2(x_2)) = \sup_{x_1 \in A_1} f_1(x_1) \sup_{x_2 \in A_2} f_2(x_2),

whenever both sides are defined.

This is an immediate consequence of the preceding corollary and Proposition 9 of no. 4.

In particular, if AA and BB are two subsets of R‾\overline{\mathbf{R}} such that the set A+BA + B of sums x+y (x∈A,y∈B)x + y \ (x \in A, y \in B) is defined, we have

(26)sup⁡(A+B)=sup⁡A+sup⁡B (26) \quad \sup (A + B) = \sup A + \sup B

if the right-hand side is defined. Again, if AA and BB are two subsets of [0,+∞][0, + \infty], we have

(27)sup⁡AB=sup⁡A⋅sup⁡B (27) \quad \sup AB = \sup A \cdot \sup B

whenever both sides are defined.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.