Bourbaki

Proposition 4

Tag 03I6, permanent. Cite it as [Bourbaki, Tag 03I6].

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

Let ff be a mapping of A into B and let (Yι)ι∈I(Y_\iota)_{\iota \in I} be a family of subsets of B. Then f−1⟨⋂ι∈IYι⟩=⋂ι∈If−1⟨Yι⟩.\overset{-1}{f}\left\langle \bigcap_{\iota \in I} Y_\iota \right\rangle = \bigcap_{\iota \in I} \overset{-1}{f}\langle Y_\iota \rangle.

To prove this, let xx be an element of ⋂ι∈If−1⟨Yι⟩\bigcap_{\iota \in I} \overset{-1}{f}\langle Y_\iota \rangle. We have f(x)∈Yιf(x) \in Y_\iota for all ι∈I\iota \in I, whence f(x)∈⋂ι∈IYιf(x) \in \bigcap_{\iota \in I} Y_\iota, and consequently

x∈f−1⟨⋂ι∈IYι⟩.x \in \overset{-1}{f}\left\langle \bigcap_{\iota \in I} Y_\iota \right\rangle.

Therefore ⋂ι∈If−1(Yι)⊂f−1⟨⋂ι∈IYι⟩\bigcap_{\iota \in I} \overset{-1}{f}(Y_\iota) \subset \overset{-1}{f}\left\langle \bigcap_{\iota \in I} Y_\iota \right\rangle; this relation, together with Proposition 3, gives the 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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Transcribed from the printed edition, 2004, Springer. Not translated.