Bourbaki

Proposition 3

Tag 03I5, permanent. Cite it as [Bourbaki, Tag 03I5].

Theory of Sets, chapter II, § 4. Union and intersection of a family of sets

Let (Xι)ι∈I(X_\iota)_{\iota \in I} be a family of subsets of a set A, and let Γ\Gamma be a correspondence between A and B. Then

Γ⟨⋃ι∈IXι⟩=⋃ι∈IΓ⟨Xι⟩,Γ⟨⋂ι∈IXι⟩⊂⋂ι∈IΓ⟨Xι⟩.\Gamma \left\langle \bigcup_{\iota \in I} X_\iota \right\rangle = \bigcup_{\iota \in I} \Gamma \langle X_\iota \rangle, \qquad \Gamma \left\langle \bigcap_{\iota \in I} X_\iota \right\rangle \subset \bigcap_{\iota \in I} \Gamma \langle X_\iota \rangle.

The relation (∃x)(x∈⋃ι∈IXι and y∈Γ(x))(\exists x)\left(x \in \bigcup_{\iota \in I} X_\iota \text{ and } y \in \Gamma(x)\right) is equivalent to

(∃x)(∃ι)(ι∈I and x∈Xι and y∈Γ(x)),(\exists x)(\exists \iota)(\iota \in I \text{ and } x \in X_\iota \text{ and } y \in \Gamma(x)),

hence to (∃ι)(ι∈I(\exists \iota)(\iota \in I and y∈Γ⟨Xι⟩)y \in \Gamma\langle X_\iota \rangle), hence is equivalent to y∈⋃ι∈IΓ⟨Xι⟩y \in \bigcup_{\iota \in I} \Gamma\langle X_\iota \rangle; this proves the first formula. As to the second formula, we have ⋂ι∈IXι⊂Xι\bigcap_{\iota \in I} X_\iota \subset X_\iota for all ι∈I\iota \in I, whence (§3, Proposition 2)

Γ⟨⋂ι∈IXι⟩⊂Γ⟨Xι⟩,\Gamma\left\langle \bigcap_{\iota \in I} X_\iota \right\rangle \subset \Gamma\langle X_\iota \rangle,

and consequently

Γ⟨⋂ι∈IXι⟩⊂⋂ι∈IΓ⟨Xι⟩.\Gamma\left\langle \bigcap_{\iota \in I} X_\iota \right\rangle \subset \bigcap_{\iota \in I} \Gamma\langle X_\iota \rangle.

¶ If Γ\Gamma is an arbitrary correspondence (and in particular an arbitrary function), the formula

Γ⟨⋂ι∈IXι⟩=⋂ι∈IΓ⟨Xι⟩\Gamma\left\langle \bigcap_{\iota \in I} X_\iota \right\rangle = \bigcap_{\iota \in I} \Gamma\langle X_\iota \rangle

is usually false.

\* For example, in the plane R2\mathbf{R}^2 the first projections of the lines y=xy = x and y=x+1y = x + 1 are equal to R\mathbf{R}, but the intersection of these lines is empty, and therefore so is the first projection of this intersection [^2]. \*

However, we have the following important result :

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

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