# Multiple of Supremum

## Theorem

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

Let $S$ be bounded above.

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

Then:

$\ds \map {\sup_{x \mathop \in S} } {z x} = z \map {\sup_{x \mathop \in S} } x$

## Proof

Let $B = \map \sup S$.

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

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

Because $z > 0$, it follows that:

$\forall x \in S: z x \le z B$

So $T$ is bounded above by $z B$.

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

We need to show that $C = z B$.

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

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 $T$:

$S = \set {z^{-1} y: y \in T}$

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:

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

So we have shown that:

$z B \le C$ and $C \le z B$

Hence the result.

$\blacksquare$