Distance from Subset of Real Numbers to Infimum

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 below such that $\xi = \inf S$.

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 $\xi = \inf S$.

Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$.

By Negative of Infimum is Supremum of Negatives:
 * $\xi = \inf S \implies -\xi = \sup S'$

Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result.