Proof
Recall that an ordered set is a set in which an order is defined. We are also given that \(B\) is not empty and that \(B\) is bounded below. By definition (1.8), this means that \(B\) has a greatest lower bound or an infimum.
Now, let \(L\) be the set of all lower bounds of \(B\). We know \(L\) is not empty since \(B\) is not empty and is bounded below. But now notice that \(L\) consists of all elements \(y \in S\) such that \(y \leq x\) for any \(x \in B\). This means that \(L\) is bounded above and that every \(x \in B\) is an upper bound of \(L\). Since \(L\) is not empty and bounded above, then \(L\) has a supremum in \(S\) by definition 1.8.
Let \(\alpha = \sup L\). We claim that \(\alpha \in L\). This means that we want to show that \(\alpha\) is a lower bound of \(B\) since by definition \(L\) contains any lower bound of \(B\). Then, note that for any \(y \in S\), if \(y < \alpha\), then \(y\) is not an upper bound of \(L\) by defintion 1.8. Since \(y\) is not an upper bound of \(L\), then \(y\) can’t be in \(B\). So this means that no element less than \(\alpha\) can be in \(B\). That is \(\alpha\) itself is a lower bound of \(B\). But then again, this means that \(\alpha \in L\) since \(L\) by definition is the set of all lower bounds of \(B\). So this establishes that \(\alpha\) is a lower bound of \(B\).
Next, we claim that \(\inf B = \alpha\). By definition 1.8, we want to show if there exists some \(\beta > \alpha\), then \(\beta\) is not a lower bound of \(B\). So suppose \(\beta > \alpha\). If that’s the case, then \(\beta\) can’t be in \(L\) since \(\alpha\) is an upper bound of \(L\). But \(L\) contains all the lower bounds of \(B\). So \(\beta\) can’t be a lower bound of \(B\). This means that \(\inf B = \alpha\). \(\ \blacksquare\)
References