Proposition 1

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

General Topology, chapter IV, § 7. Infinite sums and products of real numbers

If the families $(x_\lambda)_{\lambda \in L}$ and $(y_\mu)_{\mu \in M}$ of finite real numbers are summable in $\mathbf{R}$ , then so is the family $(x_\lambda y_\mu)_{(\lambda, \mu) \in L \times M}$ and we have

(1)

\sum_{(\lambda, \mu) \in L \times M} x_\lambda y_\mu = (\sum_{\lambda \in L} x_\lambda) (\sum_{\mu \in M} y_\mu).

Every finite subset of $L \times M$ is contained in a finite subset of the form $H \times K$ , where $H$ is a finite subset of $L$ and $K$ is a finite subset of $M$ . By hypothesis, there exists a number $a > 0$ such that $\sum_{\lambda \in H} |x_\lambda| \leq a$ and $\sum_{\mu \in K} |y_\mu| \leq a$ for all finite subsets $H$ and $K$ of $L$ and $M$ respectively; therefore

\sum_{(\lambda, \mu) \in H \times K} |x_\lambda y_\mu| = (\sum_{\lambda \in H} |x_\lambda|) (\sum_{\mu \in K} |y_\mu|) \leq a^2,

and this shows that the family

(x_\lambda y_\mu)

is summable in

\mathbf{R}

, by Theorems 1 and 3. By associativity we can write [Chapter III, § 5, no. 3, formula (2)]

\sum_{(\lambda, \mu) \in L \times M} x_\lambda y_\mu = \sum_{\lambda \in L} (\sum_{\mu \in M} x_\lambda y_\mu) = \sum_{\lambda \in L} x_\lambda (\sum_{\mu \in M} y_\mu) = (\sum_{\lambda \in L} x_\lambda) (\sum_{\mu \in M} y_\mu);

hence the result.

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

Cited by

Cites

Chapter III, § 5, no. 3