Bourbaki

Theorem 1 (Zermelo)

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

Theory of Sets, chapter III, § 2. Well-ordered sets

Every set E can be well-ordered.

Let S=P(E)−{E}\mathfrak{S} = \mathfrak{P}(\mathrm{E}) - \{\mathrm{E}\} be the set of all subsets of E other than E itself. For each X∈S\mathrm{X} \in \mathfrak{S} let p(X)=τx (x∈E−X)p(\mathrm{X}) = \tau_x \ (x \in \mathrm{E} - \mathrm{X}); since the relation X∈S\mathrm{X} \in \mathfrak{S} implies (∃x)(x∈E−X)(\exists x)(x \in \mathrm{E} - \mathrm{X}), we have p(X)∈E−Xp(\mathrm{X}) \in \mathrm{E} - \mathrm{X} (Chapter I, § 4, no. 1) and therefore p(X)∉Xp(\mathrm{X}) \notin \mathrm{X}. We may therefore apply Lemma 3, and consequently there exists a well-ordering on a subset M of E such that M∉S\mathrm{M} \notin \mathfrak{S}; but the only subset of E which does not belong to S\mathfrak{S} is E itself, and the theorem is proved.

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.