Bourbaki

Lemma 1

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

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

For each finite subset II of N\mathbf{N}, let φ(I)=sup⁡J⊂I∣∣pJ−e∣∣\varphi(I) = \sup_{J \subset I} ||p_J - e||. Then for each subset JJ of II we have

∣∣pJ−e−sJ∣∣≤φ(I)σJ. ||p_J - e - s_J|| \leq \varphi(I)\sigma_J.

The lemma is obvious if JJ is empty; we shall prove it by induction on the number of elements in JJ. Let J=K∪{j}J = K \cup \{j\}, where {j}\{j\} is strictly larger than every i∈Ki \in K. Then pJ=pK(e+aj)p_J = p_K(e + a_j) and

sJ=sK+aj, s_J = s_K + a_j,

so that

pJ−e−sJ=pK−e−sK+(pK−e)aj, p_J - e - s_J = p_K - e - s_K + (p_K - e)a_j,

and by the inductive hypothesis and the definition of φ(I)\varphi(I), we have

∣∣pJ−e−sJ∣∣≤φ(I)σK+φ(I)∣∣aj∣∣=φ(I)σJ, ||p_J - e - s_J|| \leq \varphi(I)\sigma_K + \varphi(I)||a_j|| = \varphi(I)\sigma_J,

which proves the lemma.

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

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Transcribed from the printed volume. Not translated.