Multiple of Supremum

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

Let $$S$$ be bounded above.

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

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

Proof
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 $$C$$.


 * We now need to show that $$C = zB$$.

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 $$z^{-1}$$ 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.