Bourbaki

Exercise 20, § 6

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Theory of Sets, III, 246

Bourbaki marks this one as harder than the rest.

(a) For a cardinal α\alpha to be regular (Exercise 17) it is necessary that for every cardinal β≠0\beta\neq0 we should have

αβ=α⋅∑m<αmβ. \alpha^\beta=\alpha\cdot\sum_{m<\alpha}m^\beta.

(Use Exercise 19 and consider separately the cases (i) β\beta is finite, (ii) ℵ0≤β<α\aleph_0\leq\beta<\alpha, (iii) β≥α\beta\geq\alpha; also use Exercise 3 of § 3.). The generalized continuum hypothesis implies that the above condition is also sufficient.

(b) Show that, if a cardinal α\alpha is such that αm=α\alpha^m=\alpha for every cardinal mm such that 0<m<α0<m<\alpha, then α\alpha is regular (use Exercise 3 of § 3).

(c) Show that the proposition “for every regular cardinal α\alpha and every cardinal mm such that 0<m<α0<m<\alpha, we have α=αm\alpha=\alpha^m” is equivalent to the generalized continuum hypothesis (use (a)).

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