Distance from Subset of Real Numbers to Supremum/Proof 1

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Theorem

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

Let $x \in \R$ be a real number.

Let $\map d {x, S}$ be the distance between $x$ and $S$.


Let $S$ be bounded above such that $\xi = \sup S$.

Then:

$\map d {\xi, S} = 0$


Proof

From the definition of distance:

$\forall x, y \in \R: \map d {x, y} = \size {x - y}$

Thus:

$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$


Let $\xi = \sup S$.

Then:

$\forall y \in S: \size {\xi - y} = \xi - y$

So we need to show that no $h > 0$ can be a lower bound for $T = \set {\size {\xi - y}: y \in S}$.

Aiming for a contradiction, suppose $\exists h > 0: \forall y \in S: \xi - y \ge h$.

But then:

$\forall y \in S: y \le \xi - h$

and hence $\xi - h$ is an upper bound for $T$ smaller than $\xi = \sup S$.

But this contradicts the definition of supremum, that is the smallest upper bound.

So there is no such $h > 0$.

Hence by Proof by Contradiction it follows that:

$\map d {\xi, S} = 0$.

$\blacksquare$


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