Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 4 / 1E (Analysis II)

Neither assertion holds for a general metric space. Take $X=\{0,1\}$ with $d(0,1)=1$ and center $0$. This is a discrete topology: both singleton sets are open and closed. The open ball is $U=\{0\}$, whereas the closed ball is $K=X$. Thus the interior of $K$ is $X$, strictly larger than $U$, and the closure of $U$ is $\{0\}$, strictly smaller than $K$. In general $U\subseteq \mathrm{int}K$ and $\overline{U}\subseteq K$, but the inclusions need not be equalities.

Both assertions hold for a normed vector space. The norm is continuous, so $U$ is open and $K$ is closed. If $\Vert x\Vert =1$, then $x_n=(1+1/n)x$ lies outside $K$ and converges to $x$. Hence no such boundary point belongs to the interior of $K$, giving $\mathrm{int}K=U$. Conversely, $(1-1/n)x\in U$ converges to every unit vector $x$, so $K\subseteq \overline{U}$. Closedness gives the reverse inclusion, and $\overline{U}=K$. Points already in $U$ cause no difficulty. In the zero-dimensional space $U=K=X$, so the conclusions still hold. This is the interior and closure of norm balls; radial scaling supplies what a general metric space lacks.