Bourbaki

Lemma 1

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

General Topology, chapter IX, Appendix 2. Produits infinis dans les algèbres normées

For every finite subset II of N\mathbf{N}, let φ(I)=sup⁡J⊂I∥pJ−e∥.\varphi(I) = \sup_{J \subset I} \|p_J - \mathbf{e}\|. For every subset JJ of II, one has

∥pJ−e−sJ∥≤φ(I)σJ. \|p_J - \mathbf{e} - s_J\| \leq \varphi(I) \sigma_J.
The lemma is evident if JJ is empty; let us prove it by induction on the number of elements of JJ. Let J=K∪{j}J = K \cup \{j\}, where jj is strictly upper to all the elements of KK; then pJ=pK(e+aj)p_J = p_K(\mathbf{e} + a_j) and sJ=sK+ajs_J = s_K + a_j, whence
pJ−e−sJ=(pK−e−sK)+(pK−e)aj p_J - \mathbf{e} - s_J = (p_K - \mathbf{e} - s_K) + (p_K - \mathbf{e}) a_j
and, by virtue of the induction hypothesis and the definition of φ(I)\varphi(I)
∥pJ−e−sJ∥≤φ(I)σK+φ(I)∥aj∥=φ(I)σJ \|p_J - \mathbf{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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