Bourbaki

Proposition 2

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

General Topology, chapter VII, § 3. Infinite sums in the groups $\mathbf{R}^n$

If (xi)i∈I(x_i)_{i \in I} is any finite family of points in Rn\mathbf{R}^n, then

∑i∈I∣∣xi∣∣≤2n⋅sup⁡J⊂I∣∑i∈Jxi∣. \sum_{i \in I} ||x_i|| \leq 2n \cdot \sup_{J \subset I} \left| \sum_{i \in J} x_i \right|.

For if xi=(xij)1≤j≤nx_i = (x_{ij})_{1 \leq j \leq n}, we have ∣∣xi∣∣≤∑j=1n∣xij∣||x_i|| \leq \sum_{j=1}^n |x_{ij}|, hence

∑i∈I∣∣xi∣∣≤∑j=1n(∑i∈I∣xij∣). \sum_{i \in I} ||x_i|| \leq \sum_{j=1}^n \left( \sum_{i \in I} |x_{ij}| \right).

Now ∑i∈I∣xij∣=∑i∈Ixij++∑i∈Ixij−\sum_{i \in I} |x_{ij}| = \sum_{i \in I} x_{ij}^+ + \sum_{i \in I} x_{ij}^-, and since for every subset JJ of II we have

−∑i∈Ixij−≤−∑i∈Jxij−≤∑i∈Jxij+≤∑i∈Ixij+ -\sum_{i \in I} x_{ij}^- \leq -\sum_{i \in J} x_{ij}^- \leq \sum_{i \in J} x_{ij}^+ \leq \sum_{i \in I} x_{ij}^+

it follows that

∑i∈I∣xij∣≤2⋅sup⁡J⊂I∣∑i∈Jxij∣. \sum_{i \in I} |x_{ij}| \leq 2 \cdot \sup_{J \subset I} \left| \sum_{i \in J} x_{ij} \right|.

But ∣∑i∈Jxij∣≤∣∑i∈Jxij∣\left| \sum_{i \in J} x_{ij} \right| \leq \left| \sum_{i \in J} x_{ij} \right|, hence the inequality (1).

Now, Theorem 1 is equivalent to the following proposition (since Rn\mathbf{R}^n is a complete group): the family (xi)(x_i) satisfies Cauchy’s criterion (Chapter III, § 5, no. 2, Theorem 1) if and only if the family (∣∣xi∣∣)(||x_i||) also satisfies Cauchy’s criterion. Now the triangle inequality shows that this condition is sufficient, and the inequality (1) shows that it is necessary.

Furthermore we have the inequality

∣∑ixi∣≤∑i∣∣xi∣∣ \left| \sum_i x_i \right| \leq \sum_i ||x_i||

which comes by passing to the limit from the analogous inequality for finite partial sums.

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.