# Supremum Plus Constant

## Theorem

Let $S$ be a subset of the set of real numbers $\R$.

Let $S$ be bounded above.

Let $\xi \in \R$.

Then:

$\displaystyle \sup_{x \in S} \left({x + \xi}\right) = \xi + \sup_{x \in S} \left({x}\right)$

where $\sup$ denotes supremum.

## Proof

Let $B = \sup S$.

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

Since $\forall x \in S: x \le B$ it follows that $\forall x \in S: x + \xi \le B + \xi$.

Hence $\xi + B$ is an upper bound for $T$.

If $C$ is the supremum for $T$ then $C \le \xi + B$.

On the other hand, $\forall y \in T: y \le C$.

Therefore $\forall y \in T: y - \xi \le C - \xi$.

Since $S = \left \{{y - \xi: y \in T}\right\}$ it follows that $C - \xi$ is an upper bound for $S$ and so $B \le C - \xi$.

So we have shown that $C \le \xi + B$ and $C \ge \xi + B$, hence the result.

$\blacksquare$