Take radius $r>0$. Part (a) gives one containment. For a point $x$ with $\|x-a\|\leq r$, define $x_t=a+t(x-a)$ for $0\leq t<1$. Then $\|x_t-a\|=t\|x-a\|<r$, while $x_t\to x$ as $t\uparrow1$. Hence every point of the closed ball belongs to the closure of the open ball, proving \b[equality]. Positive radius is the standard open-ball convention; a zero-radius open ball would be empty.

Solution

Past exam of the options II course of the University of Oxford / 2008 printed variant / E / E3 / d / Solution

 Solution (source code)