Let \(\mathbb{Z}_{-}\) denote the set of negative integers in the usual order. Show that a simply ordered set \(A\) fails to be well-ordered if and only if it contains a subset having the same order type as \(\mathbb{Z}_{-}\) .
Show that if \(A\) is simply ordered and every countable subset of \(A\) is well-ordered, then \(A\) is well-ordered.
Solution
- If \(A\) contains a subset order-isomorphic to \(\mathbb Z_-\), then \(A\) is not well-ordered, since that subset has no smallest element.
Conversely, suppose \(A\) is not well-ordered. Then some nonempty subset \(B\subset A\) has no smallest element. Choose \(b_1\in B\). Since \(b_1\) is not smallest in \(B\), choose \(b_2\in B\) with \(b_2<b_1\). Continuing recursively, choose \[ b_{n+1}<b_n. \] By the axiom of choice, these successive choices define a sequence \((b_n)_{n\ge1}\) in \(B\). Then \[ \cdots<b_3<b_2<b_1, \] and the map \[ -n\longmapsto b_n \] is an order isomorphism from \(\mathbb Z_-\) onto the subset \(\{b_n:n\ge1\}\subset A\).
- Suppose every countable subset of the simply ordered set \(A\) is well-ordered. If \(A\) were not well-ordered, part (a) would give a countable subset of \(A\) order-isomorphic to \(\mathbb Z_-\), which is not well-ordered. Contradiction. Hence \(A\) is well-ordered.