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

Proof
From Negative of Infimum, we have that $$-\inf_{x \in T} x = \sup_{x \in T} \left({-x}\right) \implies \inf_{x \in T} x = -\sup_{x \in T} \left({-x}\right)$$.

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

From Multiple of Supremum we have $$\sup_{x \in S} \left({zx}\right) = z \sup_{x \in S} \left({x}\right)$$.

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