Bourbaki

Proposition 5

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

Theory of Sets, chapter III, § 6. Infinite sets

The set F(E)\mathfrak{F}(\mathrm{E}) of finite subsets of an infinite set E is equipotent to E.

For each integer nn, let Fn\mathfrak{F}_n denote the set of subsets of E which have nn elements. For each X∈Fn\mathrm{X} \in \mathfrak{F}_n there exists a bijection of [1,n][1, n] onto X. Hence the cardinal of Fn\mathfrak{F}_n is at most equal to that of the set of mappings of [1,n][1, n] into E, i.e., to Card(En)=Card(E)\mathrm{Card}(\mathrm{E}^n) = \mathrm{Card}(\mathrm{E}) (no. 3, Theorem 2, Corollary 1). Therefore

Card(F(E))=∑n∈NCard(Fn)⩽Card(E)Card(N)=Card(E).\mathrm{Card}(\mathfrak{F}(\mathrm{E})) = \sum_{n \in \mathbf{N}} \mathrm{Card}(\mathfrak{F}_n) \leqslant \mathrm{Card}(\mathrm{E})\mathrm{Card}(\mathbf{N}) = \mathrm{Card}(\mathrm{E}).

On the other hand, since x→{x}x \to \{x\} is an injective mapping of E into F(E)\mathfrak{F}(\mathrm{E}), we have Card(E)⩽Card(F(E))\mathrm{Card}(\mathrm{E}) \leqslant \mathrm{Card}(\mathfrak{F}(\mathrm{E})).

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.

Cited by

Nothing.

Cites

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