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 $$S$$.

Then $$K \in \R$$ is the supremum of $$f$$ iff:


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


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

Necessary Condition
Suppose $$K \in \R$$ is the supremum of $$f: S \to \R$$.

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

Suppose $$(2)$$ were false, and:
 * $$\forall x \in S: f \left({x}\right) \le K - \epsilon$$

Then by definition, $$K - \epsilon$$ is an upper bound of $$f$$.

But by definition that means $$K \le K + \epsilon$$ 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:


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


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

From $$(1)$$ we have that $$K$$ is an upper bound of $$f$$.

Suppose that $$K$$ is not the supremum of $$f$$.

Then $$\exists M \in \R, M < K: \forall x \in S: f \left({x}\right) \le M$$

Then from, $$\exists \epsilon \in \R, \epsilon > 0: M = K - \epsilon$$.

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 supremum of $$f$$.