# 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 \(S = \mathbb{Q}\).

- (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 \(E\).
- \(E = \mathbb{Q}_{-}\), the negative rationals. \(\sup E = 0\).
- \(E = \mathbb{Q}\). \(E\) is
*unbounded above*, denoted \(\sup E = +\infty\). - \(E = \{x : x^2 < 2\}\). \(\sup E\) 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

- \(\gamma\) is an upper bound of some set \(A\) if and only if \(\sup A \leq \gamma\).
- For all \(a \in A\), \(a \leq \gamma\) if and only if \(\sup A \leq \gamma\).
- If for all \(a \in A\), \(a < \gamma\) then \(\sup A \leq \gamma\).
- If \(\gamma < \sup A\) then there exists \(a \in A\) such that \(a > \gamma\) and \(a \leq \sup A\).
- If \(A \subset B\) then \(\sup A \leq \sup B\).
- 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.

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