# Distance from Subset to Supremum

## Theorem

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

Suppose that the supremum $\sup S$ of $S$ exists.

Then:

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

where $\map d {\sup S, S}$ is the distance between $\sup S$ and $S$.

## Proof

$\map d {\sup S, S} \ge 0$

By definition of supremum:

$\forall \epsilon > 0: \exists s \in S: \map d {\sup S, s} < \epsilon$

meaning that, by nature of the infimum and the definition of $\map d {\sup S, S}$:

$\forall \epsilon > 0: \map d {\sup S, S} < \epsilon$

Together, these two observations lead to the conclusion that:

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

as desired.

$\blacksquare$