Exercise 14, § 6

(a) Show that if $\alpha \geqslant 2$ and if $\beta$ has no predecessor, then $\alpha^\beta$ is an indecomposable ordinal (cf. § 2, Exercise 16 (a)); if $\alpha$ is finite and if $\beta = \omega\gamma$, then $\alpha^\beta = \omega^\gamma$; if $\alpha$ is infinite and if $\pi$ is the greatest indecomposable ordinal $\leqslant \alpha$, then $\alpha^\beta = \pi^\beta$ (use Exercise 11).

(b) An ordinal $\delta$ is critical with respect to the functional symbol $f(\xi, \eta) = \xi\eta$ if and only if, for each $\alpha$ such that $1 < \alpha \leqslant \delta$, the equation $\delta = \alpha^\xi$ has a solution; the unique solution $\xi$ of this equation is then indecomposable (use Exercise 13 (e), together with Exercise 18 (d) of § 2). Conversely, for each $\alpha > 1$ and each indecomposable ordinal $\pi$, $\alpha^\pi$ is a critical ordinal with respect to $\xi\eta$ (use Exercise 13 (c)). Deduce that $\delta$ is a critical ordinal with respect to $\xi\eta$ if and only if $\delta$ is of the form $\omega^{\omega^\mu}$ (cf. Exercise 11 (b)).

(c) For an ordinal $\varepsilon$ to be critical with respect to the functional symbol $f(\xi, \eta) = \xi^\eta$, i.e., such that $\gamma^\varepsilon = \varepsilon$ for each $\gamma$ satisfying $2 \leqslant \gamma \leqslant \varepsilon$, it is sufficient that $2^\varepsilon = \varepsilon$. Show that the least critical ordinal $\varepsilon_0$ with respect to $\xi^\eta$ is countable (cf. Exercise 13 (d)).