Bourbaki

Corollary 3

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

Theory of Sets, chapter II, § 5. Product of a family of sets

Let (Xι)ι∈I(\mathrm{X}_\iota)_{\iota \in \mathrm{I}} and (Yι)ι∈I(\mathrm{Y}_\iota)_{\iota \in \mathrm{I}} be two families of sets having the same index set I. If Xι⊂Yι\mathrm{X}_\iota \subset \mathrm{Y}_\iota for each ι∈I\iota \in \mathrm{I}, then

∏ι∈IXι⊂∏ι∈IYι.\prod_{\iota \in \mathrm{I}} \mathrm{X}_\iota \subset \prod_{\iota \in \mathrm{I}} \mathrm{Y}_\iota.

Conversely, if ∏ι∈IXι⊂∏ι∈IYι\prod\limits_{\iota \in \mathrm{I}} \mathrm{X}_\iota \subset \prod\limits_{\iota \in \mathrm{I}} \mathrm{Y}_\iota, and if Xι≠∅\mathrm{X}_\iota \neq \emptyset for each ι∈I\iota \in \mathrm{I}, then Xι⊂Yι\mathrm{X}_\iota \subset \mathrm{Y}_\iota for each ι∈I\iota \in \mathrm{I}.

The first assertion is obvious, and the second follows from Proposition 1, Corollary 5, because we have then, for each α∈I\alpha \in \mathrm{I},

Xα=prα(∏ι∈IXι)⊂prα(∏ι∈IYι)=Yα.\mathrm{X}_\alpha = \mathrm{pr}_\alpha\Big(\prod_{\iota \in \mathrm{I}} \mathrm{X}_\iota\Big) \subset \mathrm{pr}_\alpha\Big(\prod_{\iota \in \mathrm{I}} \mathrm{Y}_\iota\Big) = \mathrm{Y}_\alpha.

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.