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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 $(X,d)$?

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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 $\operatorname{diam}(A)=\operatorname{diam}(B)$

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  • $\begingroup$ Thank you for the answer. I realized that my major problem with understanding the question was that I thought it said this condition had to hold true for ALL subets of X, rather than that we just had to find two subsets where this was true. Your answer made me realize I misread the question. Thanks again. $\endgroup$ – Jake Sep 16 '17 at 20:48
  • $\begingroup$ Glad to be of help! $\endgroup$ – Momo Sep 16 '17 at 21:00

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