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Suppose that $a,b \in R$ and $a<b$.

Now the $diam(a,b)$ = $b-a$

I am slightly confused at this point, because, by definition,

The diameter of a subset $A$ of a metric space $X$ is the $sup${$d(a,b)$|$a,b\in A$},

But in the above case $a$,$b$ do not belong to $(a,b)$ ,then why is the diameter of $(a,b)$ calculated using elements that don't belong to the set ?

johny
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    The supremum is the least upper bound. It doesn't have to be part of the set. For example, $\sup [0, 1) = 1$. – Daniel Hast Oct 09 '13 at 04:14

2 Answers2

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Because if $a,b\in\Bbb R$ and $a<b$, then

$$\operatorname{diam}(a,b)=\sup\{|x-y|:x,y\in(a,b)\}=b-a\;.$$

Added: Remember, the supremum of a set need not belong to the set: $\sup[0,1)=1$ for instance, even thought $1\notin[0,1)$.

Brian M. Scott
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I think you understand why $\mbox{diam}((a,b)) \leq b - a$. So, I'll explain why $\mbox{diam}((a,b)) \geq b - a$.

Fix $\epsilon > 0$. Then $(a + \frac{\epsilon}{2}, b-\frac{\epsilon}{2}) \subset (a,b)$. So, $\mbox{diam}(a,b) \geq b-a + \epsilon$. And since $\epsilon > 0$ was arbitrary, $\mbox{diam}((a,b)) \geq b - a$.

muffle
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