Multiple of Infimum

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

Let $T$ be bounded below.

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

Then:
 * $\ds \map {\inf_{x \mathop \in T} } {z x} = z \map {\inf_{x \mathop \in T} } x$

where $\inf$ denotes infimum.

Proof
From Negative of Infimum is Supremum of Negatives:
 * $\ds -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$

Let $S = \set {x \in \R: -x \in T}$.

From Negative of Infimum is Supremum of Negatives, $S$ is bounded above.

From Multiple of Supremum:
 * $\ds \map {\sup_{x \mathop \in S} } {z x} = z \map {\sup_{x \mathop \in S} } x$

Hence:
 * $\ds \map {\inf_{x \mathop \in T} } {z x} = -\map {\sup_{x \mathop \in T} } {-z x} = -z \map {\sup_{x \mathop \in T} } {-x} = z \map {\inf_{x \mathop \in T} } x$