Let \(S\) be a set with an order relation \(\leq\) and let \(E \subset S\). If there exists \(s \in S\) such that \(x \leq s\) for all \(x \in E\) we say that \(E\) is bounded above. The element \(s\) is called an upper bound of \(E\). Lower bounds are defined with same way with \(\geq\) in place of \(\leq.\)
Example: Let \(S = \mathbb{Q}\) and \(A = \{1, 2/3, 1/2\}\). Upper bounds for \(A\) include \(1, 10,\) and \(4/3\) since those numbers are all elements of \(S\) that are greater than every element of \(A.\).
Let \(S\) be a set with an order relation and \(E \subset S\) which is bounded above. An element \(a \in S\) is called the supremum of \(E\) if
The mathematical notation is \(a = \text{sup} E.\)
If \(E\) is bounded below an element \(a\) is called the infimum of \(E\) if
The mathematical notation is \(a = \text{inf} E.\)
Example: Let \(S = \mathbb{Q}\) and \(B = \{1 - 1/n : n \in \N\} = \{0, 1/2, 2/3, 3/4, \dots\}.\) What are \(\text{sup} B\) and \(\text{inf} B?\)
By observation we can see that \(\text{sup} B\) should be \(1\) and \(\text{inf} B\) should be \(0.\) We need to verify our intuition.
First, \(1\) is an upper bound for \(B\) since \(1 - 1/n < 1\) for every \(n \in \mathbb{N}.\) If \(c < 1\) then \(1 - c > 0.\) There must exist an \(n \in \mathbb{N}\) large enough such that \(1/n < 1-c.\) For such \(n,\) \(c < 1- 1/n\) so \(c\) is not a lower bound for \(B.\) Therefore, \(1 = \text{sup} B.\)
All of the elements of \(B\) are greater than or equal to \(0\) so \(0\) is a lower bound for \(B.\) Since \(0 \in B,\) if \(c > 0\) then \(c\) is not a lower bound. Therefore, \(0 = \text{inf} B.\)
Example: Let \(S = \mathbb{Q}\) and \(C = \{r \in S : r^2 < 2\}.\) What is \(\text{sup} C?\)
Using what we know about the real numbers, the only supremum possible for \(C\) is \(\sqrt{2}.\) However, \(S = \mathbb{Q}\) in this case and \(\sqrt{2} \notin \mathbb{Q}.\) Therefore, \(\text{sup} C\) does not exist.