Equivalence of Definitions of Supremum of Real-Valued Function

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Theorem

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

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

The following definitions of the concept of Supremum of Real-Valued Function are equivalent:

Definition 1

The supremum of $f$ on $S$ is defined by:

$\displaystyle \sup_{x \mathop \in S} f \left({x}\right) := \sup f \left[{S}\right]$

where

$\sup f \left[{S}\right]$ is the supremum in $\R$ of the image of $S$ under $f$.

Definition 2

The supremum of $f$ on $S$ is defined as $\displaystyle \sup_{x \mathop \in S} f \left({x}\right) := K \in \R$ such that:

$(1): \quad \forall x \in S: f \left({x}\right) \le K$
$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: f \left({x}\right) > K - \epsilon$


Proof

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

Aiming for a contradiction, suppose $(2)$ 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.

$\Box$


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 $f$.


Aiming for a contradiction, suppose 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.

$\blacksquare$