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 $$f$$.