Proposition 3

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

General Topology, chapter IX, Appendix 0. Infinite products in normed algebras

Let $(k_n)$ be a strictly increasing sequence of integers $\geq 0$ , such that $k_0 = 0$ ; if the infinite product with general factor $x_n$ converges, and if we put

u_n = \prod_{p=k_n}^{k_{n+1}-1} x_p,

then the infinite product whose general factor is $u_n$ is convergent and we have

\prod_{n=0}^{\infty} u_n = \prod_{n=0}^{\infty} x_n.

For the sequence of partial products of the sequence $(u_n)$ is a subsequence of the sequence of partial products of the sequence $(x_n)$ .

Finally, the same argument as was used for abelian groups (Chapter III, § 5, no. 7) shows that if a sequence $(x_n)$ in a normed algebra $\mathbf{A}$ is multiplicable, then the product whose general factor is $x_n$ is convergent, and

\prod_{n=0}^{\infty} x_n = \prod_{n \in \mathbf{N}} x_n

(which is also written as $\prod_{n=0}^{\infty} x_n$ ); the converse is of course not true (cf. Exercise 7).

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

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Nothing.

Proposition 3

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