- #1

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 413

## Homework Statement

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

## Homework Equations

Only the above, and the definition of a metric.

## The Attempt at a Solution

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

That means that it's sufficient to show that [itex]\text{diam\,}B(x,r)\geq 2r[/itex]. My idea is to show that for every [itex]\epsilon[/itex] such that [itex]0<\epsilon<2r[/itex], there exist points [itex]p,q\in B(x,r)[/itex], such that [itex]d(p,q)>2r-\epsilon[/itex]. 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 [itex]d(p,q)\leq 2r-\epsilon[/itex] 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.