Diameter of Closure of Subset is Diameter of Subset

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

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

Then:
 * $\operatorname {diam} \left({S}\right) = \operatorname {diam} \left({S^-}\right)$

where $\operatorname {diam} \left({S}\right)$ denotes the diameter of $S$, and $S^-$ denotes the closure of $S$ in $M$.

Proof
Aiming for contradiction, suppose that $\operatorname {diam} \left({S}\right) \ne \operatorname {diam} \left({S^-}\right)$.

$S \subseteq S^-$ by Subset of Metric Space is Subset of its Closure, so it then follows that:
 * $\operatorname {diam} \left({S}\right) \lt \operatorname {diam} \left({S^-}\right)$

Then there exists $x, y \in S^-$ such that $d \left({x, y}\right) \gt \operatorname {diam} \left({S}\right)$.

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

Let $d_{\infty}: \left({A \times A}\right) \times \left({A \times A}\right) \to \R$ be the metric on $A \times A$ defined as:
 * $d_{\infty} \left({\left({x, y}\right), \left({x', y'}\right)}\right) = \max \left\{{d \left({x, x'}\right), d \left({y, y'}\right)}\right\}$

From this it can be said that $\left\langle {\left({x_n, y_n}\right)} \right\rangle$ is a sequence of points in $A \times A$ that converge to $\left({x, y}\right)$.

Let $\epsilon = d \left({x, y}\right) - \operatorname {diam} \left({S}\right)$.

The Metric is Continuous, so there exists a $N \in \N_{\gt 0}$ such that:
 * $n \ge N \implies \left\vert{d \left({x_n, y_n}\right) - d \left({x, y}\right)}\right\vert \lt \epsilon$

If $d \left({x_n, y_n}\right) - d \left({x, y}\right)$ is non-negative, then:
 * $d \left({x_n, y_n}\right) \ge d \left({x, y}\right) \gt \operatorname {diam} \left({S}\right)$

which is a contradiction.

If $d \left({x_n, y_n}\right) - d \left({x, y}\right)$ is a negative number, then:

which is again, a contradiction.

So then from Proof by Contradiction it follows that $\operatorname {diam} \left({S}\right) = \operatorname {diam} \left({S^-}\right)$.

Hence the result