Question regarding diameter of subsets of a metric space 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 $X$ ?
 A: 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:$$\{\delta(X)=0\}\equiv\{X\;is\;empty\;or\;is\;composed\;of\;a\;single\;point\};\;\;\;(1)$$ 

A: 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.
