Lemma (Theorem 1.15 in the Textbook)
Suppose that \(E \subset Z\) is non-empty and bounded. Then, \(\sup E \in E\)
Proof
Suppose that \(s = \sup(E)\). Then, \(s-1\) is not a supremum. By defintion, there exists some \(x_0 \in E\) such that
$$
\begin{align*}
s - 1 < x_0 \leq s.
\end{align*}
$$
Side Note: why? because if we can't find any element \(x \in S\) such that \(x > s - \delta\), then the supremum is \(s - \delta\) and not \(s\).
We have two cases:
Case 1: If \(x_0 = s\), then \(s\) itself is an integer and we are done.
Case 2: If \(x_0 < s\), then \(x_0\) is not an upper bound of \(E\).
Side Note: since otherwise this would violate the least upper bound definition. Recall that any other upper bound must be greater than or equal to the least upper bound. So if \(x_0 < s\), then definitely it's not an upper bound.
Since \(x_0\) is not an upper bound of \(E\), then there exists another element in \(E\) such that
$$
\begin{align*}
s - 1 < x_0 < x_1 < s
\end{align*}
$$
Now, by the Order Axiom, we can add \(1-s\) to both sides to get
$$
\begin{align*}
s - 1 + 1 - s < x_0 < x_1 &< s + 1 - s \\
0 < x_0 < x_1 &< 1 \\
\end{align*}
$$
But we know that \(x_0 - x_1\) is an integer and this implies that there is an integer between \(0\) and \(1\). This is a contradiction since we don’t have an integers between \(0\) and \(1\). Therefore, we must have that \(x_0 = s. \ \blacksquare\)
References