Bourbaki

Proposition 14

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

Theory of Sets, chapter III, § 3. Equipotent sets. Cardinals

Let (aι)ι∈I(\mathfrak{a}_\iota)_{\iota \in I} and (bι)ι∈I(\mathfrak{b}_\iota)_{\iota \in I} be two families of cardinals, both indexed by the same set I, and such that aι⩾bι\mathfrak{a}_\iota \geqslant \mathfrak{b}_\iota for all ι∈I\iota \in I. Then

∑ι∈Iaι⩾∑ι∈Ibι,Pι∈I aι⩾Pι∈I bι.\sum_{\iota \in I} \mathfrak{a}_\iota \geqslant \sum_{\iota \in I} \mathfrak{b}_\iota, \qquad \mathbf{P}_{\iota \in I} \, \mathfrak{a}_\iota \geqslant \mathbf{P}_{\iota \in I} \, \mathfrak{b}_\iota.

The second inequality follows from the inclusion relations between products of sets (Chapter II, § 5, no. 4, Proposition 6, Corollary 3). As to the first inequality, if a set E is the union of a family (Aι)ι∈I(A_\iota)_{\iota \in I} of mutually disjoint subsets and if Bι⊂AιB_\iota \subset A_\iota for all ι∈I\iota \in I, then the BιB_\iota are also mutually disjoint and ⋃ιBι⊂⋃ιAι\bigcup_\iota B_\iota \subset \bigcup_\iota A_\iota (Chapter II, § 4, no. 2).

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.