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Source: Introduction to Topology and Modern Analysis by GF Simmons(Pg-69, Q.No.3)

How to show that a bounded non-empty subset of a metric space is contained within a closed sphere using the definition of a bounded set as given below?

A set is bounded if its diameter is finite (a real number less than infinity).

Also see that, given a metric space $(X,d)$, the diameter of nonempty subset $ M \subseteq X$ is defined as follows: $$\mathrm{diam}(M)=\displaystyle\sup_{x, y \in M} d(x, y)$$

Krishan
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  • When defining the diameter of a subset in a given metric space, one shouldn't forget to handle the exceptional case of the empty subset, by definition of diameter $0$. The relation with the $\sup$ does not output $0$ when applied to $\varnothing$. – ΑΘΩ Nov 17 '20 at 08:38
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    Diameter of empty set is not $0$ but $-\infty$ and diameter of singleton set is $0$. – Infinity_hunter Nov 17 '20 at 11:53
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    @ΑΘΩ the diameter of an empty set is −∞ and not 0. Diameter is the supremum of the distances between any two points of the set. the sup of ∅ is -∞. Kindly correct the edits you have made in my question. – Krishan Nov 17 '20 at 13:20
  • @Krishan and to Infinity_hunter as well: I can undo the edits if Krishan so desires, but my point is that I disagree with the $-\infty$ convention for the empty set. First of all, it is rather elementary to require diameters to be positive, and for that matter to consider the supremum operator associated to the completely ordered set $[0, \infty]$. Anyone can notice that by default the supremum of $\varnothing$ (within $\overline{\mathbb{R}}$) is $-\infty$, but the whole point is that this naive definition doesn't have much significance in the context of metric considerations. – ΑΘΩ Nov 17 '20 at 15:04
  • @Infinity_hunter Entirely agreed that singletons should be of diameter $0$, however that doesn't in any way prevent the empty set from also being conceived of as having diameter $0$. There are nuances in definitions, and if the general method of defining diameters relies on the supremum that doesn't mean we have to get tripped up by the fact that the supremum of $\varnothing$ (with respect to the completely ordered structure of $\overline{\mathbb{R}}$!) is $-\infty$. A definition can always accommodate for particular cases that are meaningful in the context at hand. – ΑΘΩ Nov 17 '20 at 15:06

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Just pick a point in the given subset, $P\in A$ and consider the ball centered at $P$ with radius $R$ equal to the diameter of $A$. Then clearly for any $Q\in A$, $dist (Q,P)\le R$ by definition of diameter.

GReyes
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  • One should not forget to handle the case of the empty subset. – ΑΘΩ Nov 17 '20 at 08:38
  • Well.. the empty subset is contained in any ball. – GReyes Nov 17 '20 at 09:32
  • We all know that, the point is that one should explicitly mention this particular case as well as long as one wishes to regard oneself as an author of carefully written, meticulous proofs. – ΑΘΩ Nov 17 '20 at 10:29
  • Dont mind but I am just looking for a stronger answer. Kindly give me another perspective if you are interested to. – Krishan Nov 17 '20 at 14:30