# Finding a condition that ensures the diameter of a proper subset and its superset are the same.

Theorem: If $(X,d)$ is a metric space and $A$ and $B$ are subsets of $X$ with $A\subseteq B$, then $\operatorname{diam}(A)\le\operatorname{diam}(B)$.

Here, $\operatorname{diam}(S)=\sup\{d(r,s):r,s\in S\}$.

"With reference to the theorem above, find a condition on a metric space $(X,d)$ that ensures that there exists subsets $A$ and $B$ of $X$ with $A\subset B$ such that $\operatorname{diam}(A)=\operatorname{diam}(B)$."

What I noticed here is that $A$ is now a proper subset of $B$. I originally was planning on having my condition be that $d$ is the discrete metric, but if $A$ is a singleton set, then $\operatorname{diam}(A)=0$ and $\operatorname{diam}(B)=1$. If $A$ is the empty set, then $\operatorname{diam}(A)=-\infty$, and $\operatorname{diam}(B)=0$ or $\operatorname{diam}(B)=1$, depending on if $B$ is singleton or not. What condition can I put on $(X,d)$?

If $X$ has at least three elements, consider $B=\{a,b,c\}\subseteq X$
WLOG we can assume $d(a,b)\ge d(a,c)$ and $d(a,b) \ge d(b,c)$
Then let $A=\{a,b\}$. Obviously $A$ is a proper subset of $B$ and $\operatorname{diam}(A)=\operatorname{diam}(B)$