Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / E / 3

The distance function $x\mapsto d(x,c)$ is continuous because the triangle inequality gives $|d(x,c)-d(y,c)|\le d(x,y)$. Hence $\overline{B}(c,r)=\{x:d(x,c)\le r\}$ is the inverse image of the closed interval $(-\infty ,r]$, so is closed. Since it contains $B(c,r)$, it also contains the closure of that open ball:

The condition is $\max (|x-1/2|,|y|)\le 1/2$. Therefore the ball is the closed squareIts four vertices are $(0,\pm 1/2)$ and $(1,\pm 1/2)$.

The induced-metric balls are intersections with the corresponding ambient balls. On the horizontal arm the distance from the centre is $|x-1/2|$; on the vertical arm it is $\max (1/2,y)$. ConsequentlyThe closure relative to $E$ of the open ball is $[0,1]\times \{0\}$. Every point $(0,y)$ with $y>0$ has positive distance $y$ from that horizontal arm, so none of those points is in the closure. In particular $(0,1/4)$ lies in the closed ball but not in the closure of the open ball, proving strict inclusion.

Let the radius be $r>0$. One inclusion follows from (a). For the reverse, if $\Vert x-c\Vert \le r$, set $x_n=c+(1-1/n)(x-c)$ for $n\ge 2$. Then $\Vert x_n-c\Vert <r$ and $x_n\to x$, so every closed-ball point is in the closure of the open ball. HenceThe positive-radius convention matters: if radius zero is allowed, the open ball is empty whereas the closed ball is $\{c\}$, so the asserted equality fails at that degenerate radius.