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.