Supremum Plus Constant
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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.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (2)$