(a) A section by any element must be countable, however, if there were the largest element $\alpha$ , the section $S_{\alpha}=S_{\Omega}-\{\alpha\}$ would be uncountable.

(b) $S_{\Omega}=S_{\alpha}\cup\{\alpha\}\cup\{x|\alpha<x\}$ is uncountable, therefore, at least one of the sets on the right must be uncountable.

(c) Every countable subset of $S_{\Omega}$ has an upper bound (Theorem 10.3). Suppose $X_{0}$ is countable, and $\alpha$ is its upper bound. Using (a) and Exercise 2(a), every $\beta\in S_{\Omega}$ has an immediate successor, which we call $s(\beta)$ . Then, the set $A=\{\alpha,s(\alpha),s^{2}(\alpha),\ldots\}$ is a countable, hence, bounded from above subset of $S_{\Omega}$ . According to Exercise 1, $A$ has the least upper bound, call it $\gamma$ . Then, $\gamma$ has no immediate predecessor. Indeed, suppose $\delta$ is an immediate predecessor of $\gamma$ , but then, $\delta<s^{n}(\alpha)\le\gamma$ for some $n\in\mathbb{Z}_{+}\cup\{0\}$ , and $\gamma=s(\delta)<s^{n+1}(\alpha)$ . We conclude that $\gamma\in X_{0}$ , and $\gamma\ge s(\alpha)>\alpha\ge\gamma\in X_{0}$ . Contradiction.

In few words, the two facts that (a) every countable subset of $S_{\Omega}$ is bounded from above, and (b) for every element of $S_{\Omega}$ there is a greater element that has no immediate predecessor, imply that $X_{0}$ cannot be bounded from above, and, hence, cannot be countable.