# Diameter of an open ball = 2*radius

Staff Emeritus
Gold Member
Not really homework, but I guess it's a "textbook-style question", so I'm putting it here.

## Homework Statement

Suppose (X,d) is a metric space, and consider an open ball $B(x,r)=\{y\in X|d(x,y)<r\}$. The diameter of a subset $E\subset X$ is defined as $\text{diam\,}E=\sup\{d(x,y)|x,y\in E\}$. I want to show that $\text{diam\,}B(x,r)=2r$.

## Homework Equations

Only the above, and the definition of a metric.

## The Attempt at a Solution

If $p,q\in B(x,r)$, we have $d(p,q)\leq d(p,x)+d(x,q)<r+r=2r$, so 2r is an upper bound for the set $\{d(x,y)|x,y\in E\}$, and diam B(x,r) is the least upper bound of that set, so $\text{diam\,}B(x,r)\leq 2r$.

That means that it's sufficient to show that $\text{diam\,}B(x,r)\geq 2r$. My idea is to show that for every $\epsilon$ such that $0<\epsilon<2r$, there exist points $p,q\in B(x,r)$, such that $d(p,q)>2r-\epsilon$. This looks like it should be really easy, and it probably is, but for some reason I can't see the solution. Maybe we can assume that $d(p,q)\leq 2r-\epsilon$ for all p,q in B(x,r), and show that this implies that B(x,r) is a subset of a smaller open ball (which of course would be a contradiction that proves the assumption false).

If the metric space is a normed vector space, this gets really easy, but I'd like to prove it (or disprove it) for an arbitrary metric space.

Office_Shredder
Staff Emeritus
Gold Member
If you just have a metric space the statement is false.

For example, X is the integers, with the normal norm d(a,b)=|a-b|. B(0,.5) is just {0} which has a diameter of 0

Staff Emeritus
D'oh. Thank you. By the way, that was the fastest useful response I ever got. 