Real Analysis

Bounds

Bounds

Let set \(E \subset S\) be ordered. If there exists \(\beta \in S\) such that for all \(x \in E\), we have \(x \leq \beta\), then \(\beta\) is an upper bound for \(E\), and \(E\) is bounded above. Similarly, if there exists \(\beta \in S\) such that for all \(x \in E\), we have \(x \geq \beta\), then \(\beta\) is a lower bound for \(E\), and \(E\) is bounded below.

For example, \(\frac{3}{2}\) is an upper bound for \(A = \{x : x^2 < 2\}\). If not, then there exists \(x > \frac{3}{2}\), which implies \(x^2 > \frac{9}{4} > 2\). So, \(x \notin A\), which is a contradiction. Thus, \(\frac{3}{2}\) is an upper bound for \(A\).

The Least Upper Bound

If there is an \(\alpha \in S\) such that (1) \(\alpha\) is an upper bound, and (2) if \(\gamma < \alpha \implies \gamma\) is not an upper bound for \(E\), then \(\alpha\) is the least upper bound for \(E\), or supremum of \(E\), denoted \(\sup E\).

For example, let \(S = \mathbb{Q}\).

1. (Finite sets have a supremum) \(E = \{\frac{1}{2}, 1, 2\}\). \(\sup E = 2\) because all elements of \(E\) is less than or equal to \(2\) and if \(\gamma < 2\) then \(\gamma\) is not an upper bound for \(E\).
2. \(E = \mathbb{Q}_{-}\), the negative rationals. \(\sup E = 0\).
3. \(E = \mathbb{Q}\). \(E\) is unbounded above, denoted \(\sup E = +\infty\).
4. \(E = \{x : x^2 < 2\}\). \(\sup E\) does not exist although \(E\) is bounded. One such bound is \(\frac{3}{2}\). . To show \(\sup A = \sup B\), one strategy is to show for all \(a \in A\), there exists \(b \in B\) such that \(a \leq b\), so \(\sup A \leq \sup B\); similarly, show \(\sup B \leq \sup A\).

The Least Upper Bound Property

A set \(S\) has the least upper bound property (or satisfies the completeness axiom) if every nonempty subset of \(S\) has an upper bound, also has least upper bound in \(S\).

Properties of Suprema

1. \(\gamma\) is an upper bound of some set \(A\) if and only if \(\sup A \leq \gamma\).
2. For all \(a \in A\), \(a \leq \gamma\) if and only if \(\sup A \leq \gamma\).
3. If for all \(a \in A\), \(a < \gamma\) then \(\sup A \leq \gamma\).
4. If \(\gamma < \sup A\) then there exists \(a \in A\) such that \(a > \gamma\) and \(a \leq \sup A\).
5. If \(A \subset B\) then \(\sup A \leq \sup B\).
6. To show \(\sup A = \sup B\), one strategy is to show for all \(a \in A\), there exists \(b \in B\) such that \(a \leq b\), so \(\sup A \leq \sup B\); similarly, show \(\sup B \leq \sup A\).

The Greatest Lower Bound

The greatest lower bound, or infinum, of a set \(A\) is denoted \(\inf A\).

The infinum exists if a set is bounded below.

Show that \(\inf A = -\sup(-A)\).

Proof. By definition, \(-A = \{-a : a \in A\}\). Let \(\alpha = \inf A\). By definition, it has the following properties.

• For all \(a \in A\), \(\alpha \leq a\).
• For all lower bound \(\gamma\) for \(A\), \(\gamma \leq \alpha\).

Since for all \(a \in A\), \(\alpha \leq a\), we have \(-a \leq -\alpha\). Thus, \(-\alpha\) is also an upper bound for \(-A\). Let \(\beta = \sup(-A)\). So, \(\beta \leq -\alpha\). Hence, \(-\beta \geq \alpha\).

For all \(a \in A\), \(-a \leq \beta\) since \(\beta\) is the least upper bound for \(-A\). Hence, for all \(a \in A\), \(a \geq -\beta\). So, \(-\beta\) is a lower bound for \(A\). Thus, \(-\beta \leq \alpha\).

Therefore, \(-\beta = \alpha\), or \(\inf A = -\sup(-A)\). QED.