# Closed set that is not $\rm proximinal$ set

Every $proximinal$ set must be $closed$, but the opposite is not true. I'm looking for such an example. A $closed$ set that is not $proximinal$ set ?

In $l^1$ consider the set $$C=\text{ cl conv }\left( \frac{n+1}ne_n \right)_{n\in\mathbb N}.$$ Then $dist(C,0)\le1$ as $\|\frac{n+1}ne_n\|_{l^1}\to1$ for $n\to\infty$.
Take $x\in C$. Then there is a sequence $(\lambda_n)$ of non-negative numbers with $\sum_{n=1}^\infty \lambda_n=1$ and $\sum_{n=1}^\infty \lambda_n \frac{n+1}ne_n =x$. Then we compute $$\|x\|_{l^1} =\sum_{n=1}^\infty \lambda_n \frac{n+1}n=1+\sum_{n=1}^\infty \lambda_n \frac{1}n.$$ This proves $\|x\|_{l^1}\ge 1$. Moreover, there is an index $k$ such that $\lambda_k>0$. Hence $\|x\|_{l^1}\ge 1+\frac1k>1$. So there is no $x\in C$ of minimal distance to the origin. And $C$ is not proximinal.