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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?

1 Answers 1

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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?2012-05-23
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    If $\overline{S}$ is compact, then $S$ has finite diamter, since compact sets have finite diameter. That's probably the easiest condition.2012-05-24
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    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.2012-05-24
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    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.2012-05-24
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    $\mathbb{R}$ is alose metric space with usual metric.Can we say that we can apply both the definition of boundedness here?And for the set having no ordering we have to use diameter concept to check boundedness?2012-05-24
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    @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.2012-05-24
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    @ThomasE. I guess by easiest I meant easiest to state.2012-05-24