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 $S$.

Then $k \in \R$ is the infimum of $f$ iff:


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


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

Necessary Condition
Suppose $k \in \R$ is the infimum of $f: S \to \R$.

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

Now let $\epsilon \in \R: \epsilon > 0$.

Suppose $(2)$ were false, and:
 * $\forall x \in S: f \left({x}\right) \ge k + \epsilon$

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

But by definition that means $k \ge k + \epsilon$ and so by Real Plus Epsilon $k > k$.

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

Sufficient Condition
Now suppose that:


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


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

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

Suppose that $k$ is not the infimum of $f$.

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

Then $\exists \epsilon \in \R, \epsilon > 0: m = k + \epsilon$.

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

This contradicts $(2)$.

So $K$ must be the infimum of $f$.