Bourbaki

Proposition 2

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

Theory of Sets, chapter II, § 2. Ordered pairs

Let A and B be two sets. The relation A×B=∅A \times B = \emptyset is equivalent to "A=∅A = \emptyset or B=∅B = \emptyset".

For the relation z∈A×Bz \in A \times B implies pr1z∈A\mathrm{pr}_1 z \in A and pr2z∈B\mathrm{pr}_2 z \in B; hence A≠∅A \neq \emptyset and B≠∅B \neq \emptyset. Conversely, the relation "x∈Ax \in A and y∈By \in B" implies (x,y)∈A×B(x, y) \in A \times B and hence A×B≠∅A \times B \neq \emptyset. In other words, the relation A×B≠∅A \times B \neq \emptyset is equivalent to "A≠∅A \neq \emptyset and B≠∅B \neq \emptyset"; hence the result.

If A, B, C are sets, we write

(A×B)×C=A×B×C;(A \times B) \times C = A \times B \times C;

an element ((x,y),z)((x, y), z) of A×B×CA \times B \times C is written (x,y,z)(x, y, z) and is called a triple. Again, if A, B, C, D are sets, we write

(A×B×C)×D=A×B×C×D;(A \times B \times C) \times D = A \times B \times C \times D;

and so on.

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.

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Cites

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