Multiple of Supremum

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Theorem

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

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

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

Let $B = \sup \left({S}\right)$.

Then by definition, $B$ is the smallest number such that $x \in S \implies x \le B$.

Let $T = \left\{{zx: x \in S}\right\}$.

Since $z > 0$, it follows that $\forall x \in S: zx \le zB$.

So $T$ is bounded above by $zB$.

By the Continuum Property, $T$ has a supremum which we will call $C$.

Since $zB$ is an upper bound for $T$, and $C$ is the smallest upper bound for $T$, it follows that $C \le zB$.

Now as $z > 0$ and is a real number, $\exists z^{-1} \in \R: z^{-1} > 0$.

So we can reverse the roles of $S$ and $T$: $S = \left\{{z^{-1}y: y \in T}\right\}$.

We know that $C$ is the smallest number such that $\forall y \in T: y \le C$

So it follows that $\forall y \in T: z^{-1} y \le z^{-1} C$.

So $z^{-1} C$ is an upper bound for $S$.

But $B$ is the smallest upper bound for $S$.

So $B \le z^{-1} C \implies zB \le C$.

So we have shown that $zB \le C$ and $C \le zB$, hence the result.

$\blacksquare$