= Solution
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 $\max(1/2,y)$. Consequently
$$
\boxed{B_E=((0,1)\times\{0\}),\qquad\overline B_E=([0,1]\times\{0\})\cup(\{0\}\times[0,1/2]).}
$$
The <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 \b[strict inclusion].
Back to article page
