Supremum Plus Constant

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

Let $$S$$ be bounded above.

Let $$\xi \in \R$$.

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

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 smallest upper bound 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.