Show that every nonempty subset of \(\mathbb{Z}\) that is bounded above has a largest element.
If \(x \notin \mathbb{Z}\), show there is exactly one \(n \in \mathbb{Z}\) such that \(n < x < n + 1\) .
If \(x - y > 1\), show there is at least one \(n \in \mathbb{Z}\) such that \(y < n < x\) .
If \(y < x\), show there is a rational number \(z\) such that \(y < z < x\) .
Solution
Let \(A\subset\mathbb Z\) be nonempty and bounded above. By the Archimedean property choose \(N\in\mathbb Z_+\) larger than an upper bound for \(A\). Then \[ B=\{N-a:a\in A\} \] is a nonempty subset of \(\mathbb Z_+\). By the well-ordering property, \(B\) has a smallest element \(b_0=N-a_0\). For every \(a\in A\), \[ N-a_0\le N-a, \] so \(a\le a_0\). Thus \(a_0\) is the largest element of \(A\).
Let \(x\notin\mathbb Z\). The set \[ A=\{m\in\mathbb Z:m<x\} \] is nonempty and bounded above, hence has a largest element \(n\) by part (a). Then \(n<x\). If \(x\ge n+1\), equality would make \(x\) an integer, while strict inequality would put \(n+1\) in \(A\), contradicting maximality of \(n\). Hence \[ n<x<n+1. \] Uniqueness follows because two distinct integers cannot both be the integer immediately below \(x\): if \(m<n\), then \(m+1\le n\), contradicting \(n<x<m+1\).
Suppose \(x-y>1\). If \(y\in\mathbb Z\), take \(n=y+1\); then \(y<n<x\). If \(y\notin\mathbb Z\), part (b) gives \(m<y<m+1\). Set \(n=m+1\). Then \[ y<n<y+1<x. \]
Let \(y<x\). Since \(x-y>0\), choose \(n\in\mathbb Z_+\) with \[ n(x-y)>1. \] Then \(nx-ny>1\), so by part (c) there is \(m\in\mathbb Z\) with \[ ny<m<nx. \] Dividing by the positive integer \(n\) gives \[ y<\frac mn<x. \] Thus \(z=m/n\in\mathbb Q\) lies strictly between \(y\) and \(x\).