Multiple of Infimum

Theorem

Let $T \subseteq \R: T \ne \varnothing$ be a non-empty subset of the set of real numbers.

Let $T$ be bounded below.

Let $z \in \R: z > 0$ be a positive real number.

Then $\displaystyle \inf_{x \in T} \left({zx}\right) = z \ \inf_{x \in T} \left({x}\right)$.

Proof

From Negative of Infimum, we have that:

$\displaystyle -\inf_{x \in T} x = \sup_{x \in T} \left({-x}\right) \implies \inf_{x \in T} x = -\sup_{x \in T} \left({-x}\right)$

Let $S = \left\{{x \in \R: -x \in T}\right\}$. From Negative of Infimum, $S$ is bounded above.

From Multiple of Supremum we have:

$\displaystyle \sup_{x \in S} \left({zx}\right) = z \ \sup_{x \in S} \left({x}\right)$

Hence:

$\displaystyle \inf_{x \in T} \left({zx}\right) = -\sup_{x \in T} \left({-zx}\right) = -z \ \sup_{x \in T} \left({-x}\right) = z \ \inf_{x \in T} \left({x}\right)$

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 2.13 \ (3)$