Suprema Lemma

Note that (i) states from the suprema is an upper bound, and then (ii) further states it's the lowest one. Thus, we can use it as a lemma:

Suprema Lemma

Suppose $s \in R$ is an upper bound for a set $A \subseteq R$ . Then $s = sup (A)$ iff, for all $ε > 0$ there is an element $a \in A$ satisfying $s - ε < a$ .

Proof
This lemma essentially says that given $s$ is an upper bound, then $s$ is the least upper bound iff any number smaller than $s$ is not an upper bound.

( $\Rightarrow$ ). Suppose $s = sup (A)$ . Let $ε > 0$ . Consider $s - ε$ . Since $s - ε < s$ and $s$ is the least upper bound, then $s - ε$ is not an upper bound for $A$ . Hence, there's some other bound $a \in A$ where $s - ε < a$ as otherwise then $s - ε$ is an upper bound for $A$ which is a contradiction.

( $\Leftarrow$ ). Assume $s$ is an upper bound with the property that no matter how $ε > 0$
is chosen that $s - ε$ is no longer an upper bound for $A$ ; namely $\forall ε > 0 \exists a \in A (s - ε < a)$ . Notice that $\forall b \in A (b < s)$ where $b$ is not an upper bound if we choose $ε = s - b$ in our given. Hence, if we assume that $b$ was an upper bound then $b \geq s$ as required (it's the opposite of the given, otherwise there's a contradiction).
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