Least Upper Bounds and Greatest Upper Bounds

Least Upper Bound (supremum)

Some $s \in R$ is the least upper bound for a set $A \subseteq R$ if it meets:

$s$ is an upper bound for $A$
if $b$ is any upper bound for $A$ then $s \leq b$
We denote the least upper bound as the supremum, or $sup (A)$ . Sometimes you'll set $s = lub (A)$ but we stick with the former.

The greatest lower bound, called the infimum for $A$ , is defined similarly and is denoted by $inf (A)$ .

Note that $A$ can have many upper bounds, but the supremum, the least upper bound, is unique. This is because using property (2) in the definition, for two suprema $s_{1}, s_{2}$ then individually $s_{1} \leq s_{2}$ and likewise $s_{2} \leq s_{1}$ , so then $s_{1} = s_{2}$ showing uniqueness.

Example

Let:

A = {\frac{1}{n} : n \in N} = {1, \frac{1}{2}, \frac{1}{3}, \dots}

$A$ is bounded above and below. An upper bound could be $3, 2, 3 / 2, . . .$ . Clearly though here $sup (A) = 1$ . We can show this using the definition.

Proof
Let's prove (i), showing that $1$ is an upper bound on $A$ . Notice that:

1 \geq \frac{1}{n} \to n \geq 1

is true since $n \in N$ , hence showing that it's a valid upper bound.

For (ii), let $b$ be another upper bound for $A$ . Since $1 \in A$ and $b$ is an upper bound for $A$ then we must have it that $1 \leq b$ , showing property (ii) exactly!
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Note that here $sup (A) \in A$ but this isn't always the case.

References

[[Abbott Real Analysis.pdf#page=28]]