Exercise 21, § 6

An infinite cardinal $\alpha$ is said to be dominant if, for each pair of cardinals $m<\alpha$, $n<\alpha$, we have $m^n<\alpha$.

(a) For $\alpha$ to be dominant it is sufficient that $2^m<\alpha$ for every cardinal $m<\alpha$.

(b) Define inductively a sequence $(a_n)$ of cardinals as follows: $a_0=\aleph_0$, $a_{n+1}=2^{a_n}$. Show that the sum $\delta$ of the sequence $(a_n)$ is a dominant cardinal. $\aleph_0$ and $\delta$ are the two smallest dominant cardinals.

(c) Show that $\delta\aleph_0=\aleph_0^\delta=2^\delta$ (note that $2^\delta\leq\delta\aleph_0$). Deduce that $\delta^{\aleph_0}=(2^{\aleph_0})^\delta$, although $\delta<2^{\delta}$ and $\aleph_0<\delta$.