Let $(X,\rho)$ be a metric space, and define

  • the sphere $S(x_0,r)\colon= \{x| \rho(x,x_0)=r\}$, and
  • the (open) ball $B(x_0,r)\colon=\{x|\rho(x,x_0)<r\}$.

Also denote

  • the interior $\mathop{Int} A$ the (inclusion-wise) greatest open set contained in $A$,
  • the closure $\mathop {Cl}A $ the (inclusion-wise) smallest closed set containing $A$, and
  • the boundary $\mathop{Fr}A \colon= \mathop{Cl}A\setminus \mathop{Int}A$.

How can the sphere be disjoint with the closure of the open ball with the same radius and centre $$ S(x_0|r) \cap \mathop{Cl} B(x_0,r) = \emptyset ?$$ Does a sphere contain the closure of the open ball with same centre and radius?

  • 1
    $\begingroup$ See related question here for the standard example. $\endgroup$ Commented Oct 24, 2022 at 1:20

1 Answer 1


Take the set $\mathbb{N}$ with the regular metric $d$ on the reals. then $$B(5,2)=\{4,5,6\} \text{ and }Cl\left(B(5,2)\right)=\{4,5,6\}$$ and $$S(5,2)=\{3,7\}$$


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