Condition Defining Infimum

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:

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

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

Proof

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:

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

$\Box$

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 $f$.

$\blacksquare$

Notes

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

Sources

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