Bourbaki

Remark 2

Tag 03S2, permanent. Cite it as [Bourbaki, Tag 03S2].

Theory of Sets, chapter II, § 3. Correspondences

The notations G(X)G(X) and Γ(X)\Gamma(X) can occasionally lead to confusion with the notation introduced later (cf. no. 4, Remark following Definition 9).

Let G be a graph. Since the relation (x,y)∈G(x, y) \in G implies y∈pr2Gy \in \mathrm{pr}_2 G, we have G⟨X⟩⊂pr2GG\langle X \rangle \subset \mathrm{pr}_2 G for every set X. Since (x,y)∈G(x, y) \in G implies x∈pr1Gx \in \mathrm{pr}_1 G, we have G⟨pr1G⟩=pr2GG\langle \mathrm{pr}_1 G \rangle = \mathrm{pr}_2 G. We have G⟨∅⟩=∅G\langle \emptyset \rangle = \emptyset, since x∉∅x \notin \emptyset is a theorem. If X⊂pr1GX \subset \mathrm{pr}_1 G and X≠∅X \neq \emptyset, we have G⟨X⟩≠∅G\langle X \rangle \neq \emptyset.

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

Nothing.

Cites

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