Let \(S_{\Omega}\) be the minimal uncountable well-ordered set.
Show that \(S_{\Omega}\) has no largest element.
Show that for every \(\alpha \in S_{\Omega}\), the subset \(\{x \mid \alpha < x\}\) is uncountable.
Let \(X_0\) be the subset of \(S_{\Omega}\) consisting of all elements \(x\) such that \(x\) has no immediate predecessor. Show that \(X_0\) is uncountable.
Solution
Recall that \(S_{\Omega}\) is uncountable and every section \(S_\alpha\) with \(\alpha\in S_\Omega\) is countable.
If \(S_\Omega\) had a largest element \(\alpha\), then \[ S_\Omega=S_\alpha\cup\{\alpha\} \] would be countable, contradiction. Hence \(S_\Omega\) has no largest element.
Fix \(\alpha\in S_\Omega\). The section \(S_\alpha\) is countable. If the tail \[ T_\alpha=\{x\in S_\Omega:\alpha<x\} \] were also countable, then \[ S_\Omega=S_\alpha\cup\{\alpha\}\cup T_\alpha \] would be countable. Hence \(T_\alpha\) is uncountable.
Let \(X_0\) be the set of points having no immediate predecessor. Suppose \(X_0\) were countable. Every countable subset of \(S_\Omega\) has an upper bound in \(S_\Omega\), so \(X_0\) would have an upper bound \(a\in S_\Omega\).
Starting at \(a\), form the successor sequence \[ a=a_0<a_1<a_2<\cdots, \] where \(a_{n+1}\) is the immediate successor of \(a_n\); this is possible by part (a) and E-MUN-10-2. The countable set \(\{a_n:n\ge0\}\) has an upper bound in \(S_\Omega\), hence, by E-MUN-10-1, a least upper bound \(\beta\in S_\Omega\).
The point \(\beta\) is not one of the \(a_n\), since \(a_{n+1}>a_n\) for every \(n\). We claim \(\beta\) has no immediate predecessor. If \(\gamma\) were an immediate predecessor of \(\beta\), then \(\gamma\) could not be an upper bound of the sequence, so \(a_n>\gamma\) for some \(n\). Then \[ \gamma<a_n<a_{n+1}<\beta, \] contradicting immediacy. Thus \(\beta\in X_0\). But \(\beta>a=a_0\), contradicting that \(a\) is an upper bound for \(X_0\). Therefore \(X_0\) is uncountable.