# 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:

$\map {\operatorname {diam} } S = \map {\operatorname {diam} } {S^-}$

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

## Proof

Aiming for a contradiction, suppose that $\map {\operatorname {diam} } S \ne \map {\operatorname {diam} } {S^-}$.

$S \subseteq S^-$ by Subset of Metric Space is Subset of its Closure, so it then follows that:

$\map {\operatorname {diam} } S < \map {\operatorname {diam} } {S^-}$

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

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

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:

$\map d {x_n, y_n} \ge \map d {x, y} > \map {\operatorname {diam} } S$

which is a contradiction.

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

 $\displaystyle \size {\map d {x_n, y_n} - \map d {x, y} }$ $=$ $\displaystyle -\map d {x_n, y_n} + \map d {x, y}$ $\displaystyle$ $<$ $\displaystyle \map d {x, y} - \map {\operatorname {diam} } S$ $\displaystyle \leadsto \ \$ $\displaystyle -\map d {x_n, y_n}$ $<$ $\displaystyle -\map {\operatorname {diam} } S$ $\displaystyle \leadsto \ \$ $\displaystyle \map d {x_n, y_n}$ $>$ $\displaystyle \map {\operatorname {diam} } S$

which is again a contradiction.

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

Hence the result.

$\blacksquare$