(related to Theorem: Supremum Property, Infimum Property)
Let \(D\) be non-empty subset of real numbers with an upper bound, say \(B_0\). We will to show that \(D\) has a supremum by constructing it (The existence of an infimum for a non-empty lower-bounded subset of real numbers can be proven in a similar way.)
We consider any \(x_0\in D\). Then we have \(x_0 < B_0\), or, equivalently, \(B_0 - x_0 > 0\). This is because \(B_0\) is an upper bound, by assumption. We now construct sequences of real numbers. * \(x_0 \le x_1 \le x_2 \le \ldots\), \(x_i\in D\) and * \(B_0 \ge B_1 \ge B_2 \ge \ldots\) of upper bounds of \(D\)
with the following property:
\[B_n-x_n\le \frac{B_0-x_0}{2^n}\quad\quad\text{for all }n\in\mathbb N\quad\quad( * )\]
in the following recursive way:
In all steps, the required properties \(x_n \le x_{n+1}\), \(B_n \ge B_{n+1}\), and \(( * )\) are all preserved. Note that the sequence \((B_n)_{n\in\mathbb N}\) is monotonically decreasing with the lower bound \(x_0\). Therefore, according to the monotone convergence theorem, it is convergent to some limit \(B\):
\[\lim_{n\to\infty}B_n=B.\]
Because convergence preserves upper bounds for sequence members, and because for any \(x\in D\), we have by construction \(x\le B\), \(B\) proves to be an upper bound of \(D\). It remains to be shown that there is no smaller upper bound of \(D\) than \(B\). Take any \(B'\) with \(B' < B\). We will show that \(B'\) is no more an upper bound of \(D\).
With a growing \(n\), the number \(2^{-n}\) becomes arbitrarily small. Therefore, we can find an \(n\) such that for any \(B' < B\) we get \[\frac{B_0-x_0}{2^n} < B - B'.\quad\quad ( * * )\]
It follows that
\[\begin{array}{rcll} x_n&\ge& B_n-\frac{B_0-x_0}{2^n}&\text{by virtue of }( * )\\ &\ge& B-\frac{B_0-x_0}{2^n}&\text{sequence }(B_n)_{n\in\mathbb N}\text{ is monotonically decreasing}\\ & > & B'&\text{by virtue of }( * * ) \end{array}\]
Therefore, there is no smaller upper bound of \(D\) than \(B\). Thus, \(B\) is a supremum.
