Equivalence of Definitions of Infimum 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 an infimum of $f$ by definition 1.

Then from the definition:


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


 * $\text{(b)}: \quad k \ge m$ for all lower bounds $m$ of $f \left({S}\right)$ in $\R$.

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

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

Aiming for a contradiction, suppose $(2)$ were false:
 * $\forall x \in S: k + \epsilon \le f \left({x}\right)$

Then by definition, $k + \epsilon$ is a lower bound of $f$.

But by definition that means $k \ge k + \epsilon$.

So by Real Plus Epsilon:
 * $k > k$

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

Thus $k$ is an infimum of $f$ by definition 2.

Definition 2 implies Definition 1
Let $k$ be an infimum of $f$ by definition 2:


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


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

From $(1)$ we have that $k$ is a lower bound of $f$.

Aiming for a contradiction, suppose that $k$ is not an infimum of $f$ by definition 1.

Then:
 * $\exists m \in \R, m > k: \forall x \in S: f \left({x}\right) \ge m$

Then:
 * $\exists \epsilon \in \R_{>0}: m = k + \epsilon$

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

This contradicts $(2)$.

Thus $k$ is an infimum of $f$ by definition 1.