Let \(A_{1}\) and \(A_{2}\) be disjoint sets, well-ordered by \(<_{1}\) and \(<_{2}\), respectively. Define an order relation on \(A_{1} \cup A_{2}\) by letting \(a < b\) either if \(a, b \in A_{1}\) and \(a <_{1} b\), or if \(a, b \in A_{2}\) and \(a <_{2} b\), or if \(a \in A_{1}\) and \(b \in A_{2}\). Show that this is a well-ordering.
Generalize (a) to an arbitrary family of pairwise disjoint well-ordered sets \(\{A_\alpha\}_{\alpha \in J}\), indexed by a well-ordered set \(J\).
Solution
Let \((J,<_J)\) be a well-ordered set, and for each \(\alpha\in J\) let \((A_\alpha,<_\alpha)\) be a well-ordered set, with \(A_\alpha\cap A_\beta=\varnothing\) for \(\alpha\ne\beta\). Put \(A=\bigcup_{\alpha\in J}A_\alpha\). For \(a\in A\) let \(\iota(a)\in J\) be the unique index with \(a\in A_{\iota(a)}\), and define a relation \(<\) on \(A\) by \[ a<b \iff \iota(a)<_J\iota(b)\quad\text{or}\quad\bigl(\iota(a)=\iota(b)\text{ and }a<_{\iota(a)}b\bigr). \] The generalization asked for in (b) is: \((A,<)\) is well-ordered. Part (a) is the case \(J=\{1,2\}\) with \(1<_J2\), where the relation above is the order defined in (a).
1. The relation \(<\) is a simple order on \(A\).
Let \(a,b\in A\) with \(\alpha=\iota(a)\) and \(\beta=\iota(b)\). If \(\alpha\ne\beta\), exactly one of \(\alpha<_J\beta\) and \(\beta<_J\alpha\) holds, so exactly one of \(a<b\) and \(b<a\) holds, and \(a\ne b\). If \(\alpha=\beta\), exactly one of \(a<_\alpha b\), \(b<_\alpha a\), \(a=b\) holds, and these are exactly \(a<b\), \(b<a\), \(a=b\). So \(<\) is comparable and nonreflexive.
For transitivity let \(a<b\) and \(b<c\) with \(\gamma=\iota(c)\). Then \(\alpha\le_J\beta\le_J\gamma\). If \(\alpha<_J\gamma\), then \(a<c\). Otherwise \(\alpha=\beta=\gamma\), so \(a<_\alpha b<_\alpha c\), hence \(a<_\alpha c\) and \(a<c\).
2. Every nonempty subset \(S\subseteq A\) has a smallest element.
The set \(J_0=\{\alpha\in J : S\cap A_\alpha\ne\varnothing\}\) is nonempty because \(S\) is, so it has a least element \(\alpha_0\). The nonempty subset \(S\cap A_{\alpha_0}\) of the well-ordered set \(A_{\alpha_0}\) has a least element \(m_0\). Let \(s\in S\) and \(\beta=\iota(s)\). Then \(\beta\in J_0\), so \(\alpha_0\le_J\beta\). If \(\alpha_0<_J\beta\), then \(m_0<s\). If \(\alpha_0=\beta\), then \(s\in S\cap A_{\alpha_0}\), so \(m_0\le_{\alpha_0}s\) and \(m_0\le s\). Hence \(m_0\) is the smallest element of \(S\).
3. Q.E.D.
Steps 1 and 2 show that \((A,<)\) is well-ordered, which is (b); taking \(J=\{1,2\}\) gives (a).