Equivalence of Definitions of Supremum of Real-Valued Function

Theorem
Let $S \subseteq \R$ be a subset of the real numbers.

Let $f: S \to \R$ be a real function on $S$.

Definition 1 implies Definition 2
Let $K \in \R$ be a supremum of $f$ by definition 1.

Then from the definition:


 * $\text{(a)}: \quad K$ is an upper bound of $f \left({x}\right)$ in $\R$.


 * $\text{(b)}: \quad K \le M$ for all upper bounds $M$ of $f \left({S}\right)$ in $\R$.

As $K$ is an upper bound it follows that:
 * $(1): \quad \forall x \in S: f \left({x}\right) \le K$

Now let $\epsilon \in \R_{>0}$.

$(2)$ were false:
 * $\forall x \in S: f \left({x}\right) \le K - \epsilon$

Then by definition, $K - \epsilon$ is an upper bound of $f$.

But by definition that means $K \le K + \epsilon$.

So by Real Plus Epsilon:
 * $K < K$

From this contradiction we conclude that:
 * $(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$

Thus $K$ is a supremum of $f$ by definition 2.

Definition 2 implies Definition 1
Let $K$ be a supremum of $f$ by definition 2:


 * $(1) \quad \forall x \in S: f \left({x}\right) \le K$


 * $(2) \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$

From $(1)$ we have that $K$ is an upper bound of $f$.

that $K$ is not a supremum of $f$ by definition 1.

Then:
 * $\exists M \in \R, M < K: \forall x \in S: f \left({x}\right) \le M$

Then:
 * $\exists \epsilon \in \R_{>0}: M = K - \epsilon$

Hence:
 * $\exists \epsilon \in \R_{>0}: \forall x \in S: f \left({x}\right) \le K - \epsilon$

This contradicts $(2)$.

Thus $K$ is a supremum of $f$ by definition 1.