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The question is :

Find a condition on a metric space$(X,d)$ that ensures that there exist subsets $A$ and $B$ of $X$ with $A \subset B$ such that $diam(A)$ = $diam(B)$.

I know that if $X$ is a metric space and $A$ and $B$ are subsets of $X$ with $A \subset B$ then

$diam(A)$ <= $diam(B)$.

If i assume that the metric $d$ on $X$ is the discrete metric then then diameters of $A$ and $B$ will be the same.

Is this good enough ? or can there be some other condition on the metric space $X$ ?

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  • $\begingroup$ HINT: >A: [0, 1] >B: [-1, 0, 1]. Yes you only need the discrete metric. Why are you trying to overcomplicate it? $\endgroup$ – Don Larynx Oct 9 '13 at 11:31
  • $\begingroup$ Reading the question literally, he just wants a sufficient condition. For example, "$(X,d)$ is the Euclidean plane$ is such a condition. Reading between the lines, maybe a more interesting condition is wanted; maybe the most general condition is wanted. $\endgroup$ – bof Oct 9 '13 at 11:58
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The condition you need is that $X$ has at least one point. If $A=\emptyset$ and $B$ is a one-point set, then $A\subset B$ and $diam(A)=diam(B)=0$.

A commenter has claimed that $diam(\emptyset)$ is undefined. I quote from Kuratowski's Topology Volume I, 1966 edition, p. 207:

III. Diameter. Continuity. Oscillation. The diameter, $\delta(X)$, of a set $X$ is the least upper bound of the distances of its points. If $\delta(X)$ is finite, the set X is said to be bounded.
The following propositions are easily proved:$$\{\delta(X)=0\}\equiv\{X\;is\;empty\;or\;is\;composed\;of\;a\;single\;point\};\;\;\;(1)$$

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  • $\begingroup$ @user103402 I think you mean sup $\emptyset=-\infty$; it is generally accepted that inf $\emptyset=+\infty$, and besides the diameter of a set is defined as a supremum, not an infimum. However, it is the supremum in the complete lattice $[0,\infty]$, not in $[-\infty,\infty]$. $\endgroup$ – bof Nov 13 '13 at 2:38
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Actually the condition you want is extremely natural. The converse is really rare. If $A$ is allowed to be empty, then the condition is just that $X$ is non-empty, as $diam(\{x\})=0=diam(\emptyset)$.

If you require $A$ to be non-empty, then the condition is $|X|\geq 3$. Indeed, let $B=\{x_1,x_2,x_3\}$ consist of three points, then $diam(B)=max\{d(x_i,x_j)\}$. Choose $A$ to be the set consisting of the two points maximising the distance.

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