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

  • $\begingroup$ 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.. $\endgroup$
    – Berci
    Mar 27 '13 at 18:51
  • $\begingroup$ I'm not sure why the set C you mentioned has to be closed $\endgroup$
    – user56714
    Mar 27 '13 at 19:44
  • $\begingroup$ @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 $e_n$. $\endgroup$
    – Jim
    Mar 28 '13 at 5:52
  • $\begingroup$ I couldn't construct such a vector ,v, yet $\endgroup$
    – user56714
    Mar 28 '13 at 10:12

Suppose $X$ infinite dimensional. Then the unit sphere $S$ is not compact (Riesz theorem), and therefore there is a sequence $x = (x_n)$ on $S$ without accumulation points. Denote by $x'$ the new sequence defined by $x_n' = (1 + \frac{1}{n})x_n$.

$x'$ and $x$ having 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 $1$ !


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