Condition Defining Supremum

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:

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

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

Proof

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:

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

$\Box$

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 $f$.

$\blacksquare$

Notes

This property can be used as a definition of the concept of supremum as applied in the context of real analysis. However, the latter definition is more general and modern.

Sources

• James M. Hyslop: Infinite Series (1942): $\S 3$