Bourbaki

Proposition 1

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

Theory of Sets, chapter III, § 1. Order relations. Ordered sets

A correspondence Γ\Gamma between E and E is an ordering on E if and only if its graph G satisfies the following conditions :

(a) G∘G=G\mathrm{G} \circ \mathrm{G} = \mathrm{G}.

(b) The set G∩G−1\mathrm{G} \cap \overset{-1}{\mathrm{G}} is the diagonal Δ\Delta of E×E\mathrm{E} \times \mathrm{E}.

For the relation ((x, y)∈G((x,\ y) \in \mathrm{G} and (y, z)∈G)⇒((x, z)∈G)(y,\ z) \in \mathrm{G}) \Rightarrow ((x,\ z) \in \mathrm{G}) can be written as G∘G⊂G\mathrm{G} \circ \mathrm{G} \subset \mathrm{G}, and the relation

((x, y)∈G and (y, x)∈G)⇔(x=y and x∈E and y∈E)((x,\ y) \in \mathrm{G} \text{ and } (y,\ x) \in \mathrm{G}) \Leftrightarrow (x = y \text{ and } x \in \mathrm{E} \text{ and } y \in \mathrm{E})

can be written as G∩G−1=Δ\mathrm{G} \cap \overset{-1}{\mathrm{G}} = \Delta. From G∩G−1=Δ\mathrm{G} \cap \overset{-1}{\mathrm{G}} = \Delta we then deduce Δ⊂G\Delta \subset \mathrm{G}, whence G=Δ∘G⊂G∘G\mathrm{G} = \Delta \circ \mathrm{G} \subset \mathrm{G} \circ \mathrm{G}; since also G∘G⊂G\mathrm{G} \circ \mathrm{G} \subset \mathrm{G}, we have

G∘G=G.\mathrm{G} \circ \mathrm{G} = \mathrm{G}.

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.

Cited by

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Cites

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Transcribed from the printed edition, 2004, Springer. Not translated.