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 a supremum of \(E\). Therefore
$$
\begin{align*}
s - 1 < x_0 < s \quad\quad\quad (1)
\end{align*}
$$
Recall that \(x_0 \in E\). Hence, it is an integer. We now claim that it is impossible to have \(x_0\) be an integer while satisfying the above inequality. To see this, define
…………. TODO … removed the earlier proof since it didn’t make sense
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