# Distance between a point and closed set in finite dimensional space

Let $X$ be a linear normed space. I need to prove that $X$ is finite dimensional normed space if and only if for every non empty closed set $C$ contained in $X$ and for every $x$ in $X$ the distance $d(x,C)$ is achieved in specific $c$. I know how to prove the direction which assumes $X$ is finite dimensional (use Riesz lemma) but I dont know what to do in the other direction. thanks

• In an infinite dimensional space with normed basis $e_1,e_2,..,e_n,..$, I guess, their set $C:=\{e_1,e_2,..,e_n,..\}$ is closed. Probably we can construct a vector $v\in X$ such that $d(v,e_n)$ is strictly decreasing.. Mar 27, 2013 at 18:51
• I'm not sure why the set C you mentioned has to be closed Mar 27, 2013 at 19:44
• @Berci: As each vector in $X$ must be a linear combination of finitely many of the basis vectors, $d(v, e_n)$ takes on only finitely many different values. So in this case $d(v, C)$ should be attained by one of the $e_n$.
– Jim
Mar 28, 2013 at 5:52
• I couldn't construct such a vector ,v, yet Mar 28, 2013 at 10:12

Suppose $$X$$ infinite dimensional. Then the unit sphere $$S$$ is not compact (Riesz theorem), and therefore there is a sequence $$x = (x_n)_n$$ on $$S$$ without accumulation points. Denote by $$x'$$ the new sequence defined by $$x_n' = (1 + \frac{1}{n})x_n$$.
Since $$x'$$ and $$x$$ have the same accumulation points, $$x'$$ doesn't have any. So the set $$C$$ of the values of $$x'$$ is closed.
Now $$d(0, C) = 1$$, and there is no point in $$C$$ of norm $$1$$.