# 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.

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