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:
 * $\ds \map {\sup_{x \mathop \in S} } {x + \xi} = \xi + \map {\sup_{x \mathop \in S} } x$

where $\sup$ denotes supremum.

Proof
Let $B = \sup S$.

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

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 $S = \set {y - \xi: y \in T}$ 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.