Bourbaki

Proposition 6

Tag 06WV, permanent. Cite it as [Bourbaki, Tag 06WV].

Algebra, chapter VI, § 1. Ordered groups. Divisibility

— Let (xi)i∈A,(yβ)β∈B(x_i)_{i \in A}, (y_\beta)_{\beta \in B} be two families of elements of an ordered group G, each having a supremum. Then the family (xα+yβ)(α,β)∈A×B(x_\alpha + y_\beta)_{(\alpha, \beta) \in A \times B} has a supremum, and

sup⁡(α,β)∈A×B(xα+yβ)=sup⁡α∈Axα+sup⁡β∈Byβ. \sup_{(\alpha, \beta) \in A \times B} (x_\alpha + y_\beta) = \sup_{\alpha \in A} x_\alpha + \sup_{\beta \in B} y_\beta.
Indeed, from xα+yβ≤zx_\alpha + y_\beta \leq z for all α\alpha and β\beta, we deduce sup⁡(xi)+sup⁡(yβ)≤z\sup(x_i) + \sup(y_\beta) \leq z for all β\beta, and hence sup⁡(xi)+sup⁡(yβ)≤z\sup(x_i) + \sup(y_\beta) \leq z.

1 When A is the ring of integers (resp. the polynomial ring in one indeterminate with coefficients in a field), these definitions coincide with those of I, p. 112 (resp. IV, p. 12, Def. 1).

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

Cited by

Nothing.

Cites

Transcribed from the printed edition, 2003, Springer. Not translated.