Is it the case that if $B(x, r)$ is a ball in a metric space $X$, where $x\in X$, then $B(x, r)=B \bigg(x, \frac{\text{diam}(B(x, r))}{2}\bigg) ?$ I believe this is correct but I want to check, thanks. ($\text{diam}$ is the diameter)
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$x$ is from any metric space? – zkutch Aug 30 '20 at 11:52
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@zkutch yes $x$ is in the metric space. – Loli Aug 30 '20 at 11:54
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@zkutch i have edited the question to clarify. – Loli Aug 30 '20 at 11:56
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This is not true.
On the metric space $X = \lbrace 0, 2 \rbrace$ with the metric induced by the usual metric on $\mathbb{R}$, one has $$B(0, 3) = \lbrace 0, 2 \rbrace$$
but $\mathrm{diam}(B(0, 3))=2$, and $$B(0,1)= \lbrace 0 \rbrace$$
TheSilverDoe
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Thank you. then I think instead this is true $B(x, r)=B (x, \text{diam}(B(x, r)))$ ? – Loli Aug 30 '20 at 12:06
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No :) In $\mathbb{R}$ with the usual metric, the diameter of $B(0,1)$ is $2$, and $B(0,1) \neq B(0,2)$... – TheSilverDoe Aug 30 '20 at 12:08
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We have $\operatorname{diam}(B(x,r)) \le 2r$. In general, it is not equality.
In extreme cases, we could have $B(x,r) = \{x\}$ with $\operatorname{diam}(B(x,r))=0$.
In an Ultrametric space we have $\operatorname{diam}(B(x,r)) \le r$.
In a normed vector space (over the real or complex numbers) we do have $\operatorname{diam}(B(x,r)) = 2r$.
GEdgar
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