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Not really homework, but I guess it's a "textbook-style question", so I'm putting it here.
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.
Only the above, and the definition of a metric.
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.
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.
