Past exam of the options II course of the University of Oxford / 2008 / E / E3

For a fixed centre $a$, the triangle inequality gives $|d(a,x)-d(a,y)|\le d(x,y)$, so the distance-to-centre function is continuous. Its preimage of $(-\infty ,r]$ is the closed ball, hence is closed. Since that closed set contains the corresponding open ball, it contains its closure:Here the bar on the right denotes a closed ball, while the bar on the left denotes closure.

The condition $\max (|x-1/2|,|y|)\le 1/2$ gives the closed squareIts four edges and corners are included, as shown in the left panel of the figure below.

Intersect the ambient balls with $E$. All points on the vertical arm have distance at least $1/2$ from the centre, so the open ball isThe closed ball also contains the endpoints of that segment and the lower half of the vertical arm:The closure of the open ball in $E$ is just $[0,1]\times \{0\}$. Both endpoints are limits of points in the open segment, whereas a vertical point of positive height has a small enough neighbourhood disjoint from it. Thus every point $(0,y)$ with $0<y\le 1/2$ witnesses strict containment.

Take radius $r>0$. Part (a) gives one containment. For a point $x$ with $\Vert x-a\Vert \le r$, define $x_t=a+t(x-a)$ for $0\le t<1$. Then $\Vert x_t-a\Vert =t\Vert x-a\Vert <r$, while $x_t\to x$ as $t\uparrow 1$. Hence every point of the closed ball belongs to the closure of the open ball, proving equality. Positive radius is the standard open-ball convention; a zero-radius open ball would be empty.