Equality of diameters in metric space

Question

Does it hold that in any metric space: $diam S\left ( p,r \right )= diam \tilde{S}\left ( p,r \right )$?

My solution:

When we consider trivial metric, where:

d(x,y)=0, if x=y
d(x,y)=1, if x $\neq $ y,

then it does, because diameter is the supremum from all of the distances, thus it is equal to 1 and it does not matter if we have open or close ball.

Is that correct?

Is $\tilde S$ the closed ball? – csch2 Mar 18 '20 at 19:59 — csch2, Mar 18 '20 at 19:59

score 2 · Answer 1 · answered Mar 18 '20 at 20:05

2

Here is a counterexample. Consider the metric space $(\mathbb{E},d)$ with $\mathbb{E}=\{(0,0)\}\cup\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ and let $d$ be the standard metric on $\mathbb{R}^2$. Consider $S\big((0,0),1\big)$ and $\tilde S\big((0,0),1\big)$. $S\big((0,0),1)=\{(0,0)\}$, which has diameter zero, but $\operatorname{diam}\tilde S\big((0,0),1\big)=2$.

answered Mar 18 '20 at 20:05

csch2

4,557

Thank you so much – Martin N. Mar 18 '20 at 20:16

score 1 · Accepted Answer · answered Mar 18 '20 at 20:09

It is not true and the discrete metric space $M$ (with $\vert M \vert >1$) is actually the counter example. There we have $S(x,1) = \{x\}$ since $x$ itself is the only point having distance stricly smaller than $1$. But $\overline{S}(x,1) = M$ since for every $y \in M$ we have $d(x,y) = 1 \leq 1$.

Thus we have $$\text{diam}S(x,1) = \sup_{y \in S(x,1)} d(x,y) = d(x,x) = 0$$ but $$\text{diam}\overline{S}(x,1) = \sup_{y \in M} d(x,y) = 1 ~~.$$

All of these examples rely on the fact that in a general metric space $\overline{S}(x,1)$ need not be equal to $\overline{S(x,1)}$. If that is the case, then your assertion is true since for any set $\text{diam}(A) = \text{diam}(\overline{A})$.

For example, for metric spaces where the metric is induced by a norm, the formula $\overline{S}(x,1) = \overline{S(x,1)}$ is true.

Equality of diameters in metric space

2 Answers2