# A basic question on diameter of a metric space

A set S of real numbers is bounded if it has both upper and lower bounds. Therefore, a set of real numbers is bounded if it is contained in a finite interval.

While in a metric space a non-empty subset $S$ of metric space $X$ is said to be bounded set if its diameter is finite.

My question is: are both the definitions of boundedness are equivalent? Is it possible to determine diameter of every subset?

Is there any other criterion also to determine whether given subset of a metric space is bounded or not?

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## 1 Answer

The diameter of a nonempty subset $S\subseteq X$ is defined to be $$\sup_{x,y\in S}d(x,y),$$ which is always defined (though it may be infinite). Thus you can always determine the diameter of a nonempty subset.

A nonempty subset $S\subseteq X$ having finite diameter is equivalent to $S$ being contained in some ball $B(x,r)$ for some $x\in X$ and some $r>0$. To see this, first suppose that $S$ has finite diameter $\Delta<\infty$. Pick any $x\in S$. Then $S\subseteq B(x,2\Delta)$, since $d(x,y)\leq \Delta$ for every $y\in S$. On the other hand, suppose that $S$ is contained in some ball $B(x,r)$, where $x\in X$ and $r>0$. Then for any two points $y,z\in S$, one has $d(y,z)\leq d(y,x) + d(x,z)\leq 2r$, so the diameter of $S$ is $\leq 2r$. I hope that answered your question!

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Is there any other criterion also to determine whether given subset of a metric space is bounded or not? – srijan May 23 '12 at 19:30
If $\overline{S}$ is compact, then $S$ has finite diamter, since compact sets have finite diameter. That's probably the easiest condition. – froggie May 24 '12 at 1:03
In my second question i asked why to find diameter of a set to check boundedness in a metric space? Can't we use same crietrion as we do in real line:Finding lower and upper bound of a given set. – srijan May 24 '12 at 2:01
For a general metric space, what do upper bound and lower bound mean? In $\mathbb{R}$, there is an ordering, which allows you to talk about upper and lower bounds. – froggie May 24 '12 at 2:03
@froggie: In my opinion that is a rather strict requirement, and I'm not quite sure what you mean by the 'easiest'. It is sufficient for sure but it is far away from being necessary: a set having finite diameter is very far away from its closure being compact. A set with a compact closure is called relatively compact. This would in a sense identify relative compact sets with bounded sets, while the first property is very strong and the latter very weak. – T. Eskin May 24 '12 at 7:24

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