Chapter 2 Analysis

2.1 Upper bounds and suprema

This section is drawn from Lay (2005).

Theorem 2.1 Let \(x,y\in\mathbb{R}\) such that \(x\leq y+\epsilon\) for every \(\epsilon>0\). Then \(x\leq y\).

Definition 2.1 Let \(\mathcal{S}\) be a subset of \(\mathbb{R}\). If there exists a real number \(m\) such that \(m\geq s\) for all \(s\in \mathcal{S}\), then \(m\) is called an upper bound for \(\mathcal{S}\), and we say that \(\mathcal{S}\) is bounded above. If \(m\leq s\) for all \(s\in \mathcal{S}\), then \(m\) is a lower bound for \(\mathcal{S}\) and \(\mathcal{S}\) is bounded below. The set \(\mathcal{S}\) is said to be bounded if it is bounded above and bounded below.

If an upper bound \(m\) for \(\mathcal{S}\) is a member of \(\mathcal{S}\), then \(m\) is called the maximum (or largest element) of \(\mathcal{S}\), and we write \(m=\max \mathcal{S}\).

Similarly, if a lower bound of \(\mathcal{S}\) is a member of \(\mathcal{S}\), then it is called the minimum (or least element) of \(\mathcal{S}\), denoted by \(\min \mathcal{S}\).

A set may have upper or lower bounds, or it may have neither. If \(m\) is an upper bound for \(\mathcal{S}\), then any number greater than \(m\) is also an upper bound. While a set may have many upper and lower bounds, if it has a maximum or a minimum, then those values are unique. Thus we speak of an upper bound and the maximum.

Definition 2.2 Let \(\mathcal{S}\) be a nonempty subset of \(\mathbb{R}\). If \(\mathcal{S}\) is bounded above, then the least upper bound of \(\mathcal{S}\) is called its supremum and is denoted by \(\sup \mathcal{S}\). Thus \(m=\sup \mathcal{S}\) iff

  1. \(m\geq s\), for all \(s\in \mathcal{S}\), and
  2. if \(m'<m\), then there exists \(s'\in \mathcal{S}\) such that \(s'>m'\).
If \(\mathcal{S}\) is bounded below, then the greatest lower bound of \(\mathcal{S}\) is called its infimum and is denoted by \(\inf \mathcal{S}\).
Definition 2.3 (completeness axiom) Every nonempty subset \(\mathcal{S}\) of \(\mathbb{R}\) that is bounded above has a least upper bound. That is, \(\sup \mathcal{S}\) exists and is a real number.
Theorem 2.2 Given nonempty subsets \(\mathcal{A}\) and \(\mathcal{B}\) of \(\mathbb{R}\), let \(\mathcal{C}\) denote the set \[ \mathcal{C}=\left\{ x+y:x\in \mathcal{A}\text{ and }y\in \mathcal{B}\right\} . \] If \(\mathcal{A}\) and \(\mathcal{B}\) have suprema, then \(\mathcal{C}\) has a supremum and \(\sup \mathcal{C}=\sup \mathcal{A}+\sup \mathcal{B}\).

Proof. Let \(\sup \mathcal{A}=a\) and \(\sup \mathcal{B}=b\). If \(z\in \mathcal{C}\), then \(z=x+y\) for some \(x\in \mathcal{A}\) and \(y\in \mathcal{B}\). Thus \(z=x+y\leq a+b\), so \(a+b\) is an upper bound for \(\mathcal{C}\). By the completeness axiom, \(\mathcal{C}\) has at least an upper bound, say \(\sup \mathcal{C}=c\). We must show that \(c=a+b\). Since \(c\) is the least upper bound for \(\mathcal{C}\), we have \(c\leq a+b\).

To see that \(a+b\leq c\), choose any \(\epsilon>0\). Since \(a=\sup \mathcal{A}\), \(a-\epsilon\) is not an upper bound for \(\mathcal{A}\), and there must exist \(x\in \mathcal{A}\) such that \(a-\epsilon<x\). Similarly, since \(b=\sup \mathcal{B}\), there exists \(y\in \mathcal{B}\) such that \(b-\epsilon<y\). Combining these inequalities, we have \[ a+b-2\epsilon<x+y\leq c. \] That is, \(a+b<c+2\epsilon\) for every \(\epsilon>0\). Thus, by Theorem 2.1, \(a+b\leq c\). Finally, since \(c\leq a+b\) and \(c\geq a+b\), we conclude that \(c=a+b\).

References

Lay, S.R. 2005. Analysis: With an Introduction to Proof. Pearson Prentice Hall. https://books.google.com/books?id=k4k\_AQAAIAAJ.