= Solution
\b[Both assertions hold for a <normed vector space>.] The <norm> is continuous, so $U$ is open and $K$ is closed. If $\|x\|=1$, 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 $\operatorname{int}K=U$. 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.
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