# In a normed space, characterization of the sets $F$ such that $d(x,F)$ is achieved for all $x$?

Let $(E,\|.\|)$ be a real normed vector space and $d$ be the distance associated to the norm. I am wondering if there exists a characterization of the subsets $F$ of $E$ such that for all $x\in E$, there exists $y\in F$ satisfying $d(x,y)=d(x,F)$. If $E$ is finite dimensional, these sets are exactly the closed sets, but in the infinite dimensional case, the closedness is only necessary. Are there additional constraints which give $F$ this property?

• Compactness is sufficient. If the norm is nice enough, convexity (in addition to closedness, of course) suffices. – Daniel Fischer Oct 4 '13 at 15:35

Let $M$ be a subspace of a normed linear space $E$, and $M^{\perp}$ denote its annihilator in the dual space $E^{\ast}$. Then every linear functional on $M$ has a unique norm-preserving extension to $E$ iff for every $f\in E^{\ast}$, there is a unique best approximation $g\in M^{\perp}$ such that $\|f-g\| = d(f,M^{\perp})$