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 $S$.

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:

$\ds \sup_{x \mathop \in S} \map f x := \sup f \sqbrk S$

where

$\sup f \sqbrk S$ is the supremum in $\R$ of the image of $S$ under $f$.

Definition 2

The supremum of $f$ on $S$ is defined as $\ds \sup_{x \mathop \in S} \map f x := K \in \R$ such that:

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

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 $\map f x$ in $\R$.
$\text{(b)}: \quad K \le M$ for all upper bounds $M$ of $f \sqbrk S$ in $\R$.

As $K$ is an upper bound it follows that:

$(1): \quad \forall x \in S: \map f x \le K$

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

Aiming for a contradiction, suppose $(2)$ were false:

$\forall x \in S: \map f x \le K - \epsilon$

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:

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

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:

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

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:

$\exists M \in \R, M < K: \forall x \in S: \map f x \le M$

Then:

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

Hence:

$\exists \epsilon \in \R_{>0}: \forall x \in S: \map f x \le K - \epsilon$

This contradicts $(2)$.

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

$\blacksquare$