Let $H$ be a Hilbert space, and let $K$ be a non-empty closed subset of $H$. Fix a point $x\in H$. If $K$ is convex, the Hilbert projection theorem gives us a unique distance minimizer $y\in K$, in the sense that $\|y-x\|\le\|y'-x\|$ for all $y'\in K$. If we do not need uniqueness, we can replace the convexity assumption with compactness. In particular, if $H=\mathbf R^d$, any closed $K$ will have a distance minimizer (since if we set $D:=\inf_{y'\in K}\|y'-x\|<\infty$, we may consider the compact set $\overline{B(x,D+1)}\cap K$).
I am interested in situations where no distance minimizer exists. For example, if $H=\mathbf R$, $K=(0,1)$, and $x=0$, so that $K$ is convex but not closed, then no distance minimizer exists, as all points of $K$ are at a positive distance away from $x$, whereas $\inf_{y'\in(0,1)}\|y'-x\|=0$.
How do I construct a non-convex closed subset $K$ of a Hilbert space $H$ for which no distance minimizer exists? As discussed earlier, $H$ cannot be $\mathbf R^d$ or $\mathbf C^d$. I tried working with $\ell^2(\mathbf Z)$ but did not get far. (I would prefer just a hint, rather than a full solution.)
Thank you.
Best Answer
You can take $$ K = \{ (1+1/n) e_n \mid n \in \mathbb N\},$$ where $e_n$ are the canonical unit sequences. Then, $K$ is closed, the distance of $0$ to $K$ is $1$, but this distance is not attained.