Diameter of Closure of Subset is Diameter of Subset

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $S \subseteq A$ be bounded in $M$.

Then:
 * $\map \diam S = \map \diam {S^-}$

where $\map \diam S$ denotes the diameter of $S$, and $S^-$ denotes the closure of $S$ in $M$.

Proof
that $\map \diam S \ne \map \diam {S^-}$.

$S \subseteq S^-$ by Subset of Metric Space is Subset of its Closure, so it then follows that:
 * $\map \diam S < \map \diam {S^-}$

Then there exists $x, y \in S^-$ such that $\map d {x, y} > \map \diam S$.

By Point in Closure of Subset of Metric Space iff Limit of Sequence there exists sequences $\sequence {x_n}$ and $\sequence {y_n}$ that converge to $x$ and $y$ respectively.

Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined as:
 * $\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$

From this it can be said that $\sequence {\tuple {x_n, y_n} }$ is a sequence of points in $A \times A$ that converge to $\tuple {x, y}$.

Let $\epsilon = \map d {x, y} - \map \diam S$.

From Distance Function of Metric Space is Continuous, there exists a $N \in \N_{> 0}$ such that:
 * $n \ge N \implies \size {\map d {x_n, y_n} - \map d {x, y} } < \epsilon$

If $\map d {x_n, y_n} - \map d {x, y}$ is non-negative, then:
 * $\map d {x_n, y_n} \ge \map d {x, y} > \map \diam S$

which is a contradiction.

If $\map d {x_n, y_n} - \map d {x, y}$ is a negative number, then:

which is again a contradiction.

From Proof by Contradiction it follows that $\map \diam S = \map \diam {S^-}$.

Hence the result.