91Ó°ÊÓ

Firstly, it's necessary to understand that the supremum of set \(S \subseteq \mathbb{R}\) is the smallest real number that is greater than or equal to every number in \(S\). So, to show that \(S\) contains its supremum, the proof must demonstrate that the largest number in \(S\) is also the smallest real number greater than or equal to all numbers in \(S\).

In this step, it's important to check the statement in the base case. To do this, start with the smallest possible set size, which is 1. For a set \(S\) with a single element, that element is both the largest number in the set and the smallest number greater than or equal to all other numbers. Therefore, the supremum is in \(S\), and the base case is proved.

Next, it's time to proceed to the inductive step. The goal here is to assume that the claim holds for some particular positive integer \(k\), and then to prove that it must also hold for \(k+1\). In this scenario, if \(S\) is a nonempty finite subset of \(\mathbb{R}\) with \(k+1\) elements, it can be written as \(S = A \cup \{b\}\), where \(A\) is a set with \(k\) elements and \(b \notin A\). If \(b\) is the maximum of \(S\), then \(b\) is the supremum and is in \(S\). If \(b\) is not the maximum of \(S\), then the supremum of \(S\) is the same as the supremum of \(A\), which by the induction hypothesis is in \(A\) and consequently is in \(S\). So, for either case, the supremum of \(S\) is in \(S\). The hypothesis hence holds for \(k+1\) if it holds for \(k\). Consequently, by induction, the claim is proved.