Proposition 3

Tag 03LV, permanent. Cite it as [Bourbaki, Tag 03LV].

Theory of Sets, chapter III, § 5. Properties of integers

Let $(a_i)_{i \in I}$ and $(b)_{i \in I}$ be two finite families of integers such that $a_i \leqslant b_i$ for each $i \in I$ and $a_i < b_i$ for at least one index $i$ . Then

\sum_{i \in I} a_i < \sum_{i \in I} b_i.

If also $b_i > 0$ for each $i \in I$ , then

\prod_{i \in I} a_i < \prod_{i \in I} b_i.

Let $j$ be an index such that $a_j < b_j$ , and let $J = I - \{j\}$ . Then

b_j = a_j + c_j

with $c_j > 0$ (Proposition 2), and therefore (§ 3, no. 6, Proposition 14)

\sum_{i \in I} b_i = a_j + c_j + \sum_{i \in J} b_i \geqslant c_j + a_j + \sum_{i \in J} a_i = c_j + \sum_{i \in I} a_i.

Since $c_j > 0$ , the first part of the Proposition follows from Proposition 2. Likewise

\prod_{i \in I} b_i = (a_j + c_j) \prod_{i \in J} b_i = a_j . \prod_{i \in J} b_i + c_j . \prod_{i \in J} b_i \geqslant \prod_{i \in I} a_i + c_j . \prod_{i \in J} b_i.

Since $c_j$ and all the $b_i$ are $\neq 0$ , the product $c_j . \prod_{i \in J} b_i$ is $\neq 0$ (§ 3, no. 4, Proposition 7); hence the second part of the Proposition.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Proposition 3

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