Bourbaki

Remark

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

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

Let E be a set, let A be a subset of E, let gg be a mapping of A into E, and let aa be an element of A. Take S{u}S\{u\} to be the term g(u)g(u). Criterion C63 is applicable and proves the existence of a mapping ff of N\mathbf{N} onto a set V such that f(0)=af(0) = a and f(n+1)=g(f(n))f(n + 1) = g(f(n)) for every integer nn. It may happen that V⊂A\mathrm{V} \subset \mathrm{A}; if not, let pp be the largest integer such that f([0,p])⊂Af([0, p]) \subset \mathrm{A}. Then f(p+1)=g(p)∉Af(p + 1) = g(p) \notin \mathrm{A}, and g(g(p))g(g(p)) is a term about which nothing can be said. Hence in this case ff is considered to be defined only on the interval [0,p+1][0, p + 1] ("restricted induction").

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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Cites

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