Bourbaki

Proposition 6

Tag 03R3, permanent. Cite it as [Bourbaki, Tag 03R3].

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

Let (aι)ι∈I(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}} be a family of cardinals, and let J\mathrm{J} (resp. K\mathrm{K}) be a subset of I\mathrm{I} such that aι=0\mathfrak{a}_\iota = 0 for all ι∉J\iota \notin \mathrm{J} (resp. aι=1\mathfrak{a}_\iota = 1 for all ι∉K\iota \notin \mathrm{K}). Then

∑ι∈Iaι=∑ι∈Jaι(resp. Pι∈I aι=Pι∈K aι).\sum_{\iota \in \mathrm{I}} \mathfrak{a}_\iota = \sum_{\iota \in \mathrm{J}} \mathfrak{a}_\iota \qquad \left(\text{resp. } \mathbf{P}_{\iota \in \mathrm{I}}\, \mathfrak{a}_\iota = \mathbf{P}_{\iota \in \mathrm{K}}\, \mathfrak{a}_\iota\right).

The proposition is obvious as regards the sum, for the sum SI\mathrm{S}_\mathrm{I} of the family of sets (aι)ι∈I(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}} is equipotent to the union of the sum SJ\mathrm{S}_\mathrm{J} of the family (aι)ι∈I(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}} and the empty set, and hence equipotent to SJ\mathrm{S}_\mathrm{J}. The assertion concerning products follows from the fact that the projection prK\mathrm{pr}_\mathrm{K} of the product set ∏ι∈Iaι\prod_{\iota \in \mathrm{I}} \mathfrak{a}_\iota onto the partial product ∏ι∈Kaι\prod_{\iota \in \mathrm{K}} \mathfrak{a}_\iota is a bijection (Chapter II, § 5, no. 5, Remark 1).

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.