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Let $A \subset X$ where $X$ is a metric space. by definition diam$(A) = \sup\{ d(x,y), x,y \in A\}$.

if $A$ is non empty and has zero diameter, can I conclude that $A$ is a singleton?

i reason as follows: there is at least one $a \in A$. If there is $b \in A$ such that $a \neq b$, then $d(a,b) > 0$ so diam$(A) > 0$. so contradiction.

is this correct? just making sure. thanks

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    $\begingroup$ Yes, it’s correct. $\endgroup$ – Brian M. Scott Apr 9 '16 at 23:18

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