# Best approximation of a function out of a closed subset

I'm studying approximation theory and I saw this exercise on Rivlin book an introduction to the approximation of functions:

Prove that if $V$ is a normed linear space, $W$ a finite-dimensional subspace of $V$, and $U$ a closed subset of $W$, then, given $v\in V$, there exists $u^*\in U$ such that $\Vert v-u^*\Vert\le\Vert v-u\Vert$ for all $u\in U$.

my approach is as follows:

$dist(v,U)=inf\Vert v-u\Vert_{u\in U}$
by definition of infimum there exists a sequence $[u_i]$ in $U$ such that $lim_{i\to \infty}\Vert v-u_i \Vert=dist(v,U)$ since $U$ is closed it contains all it's limit points and $u^*=dist(v,U)$ belongs to $U$.

so I didn't use the assumption that W is a finite-dimensional subspace.
Am I missing something here? if yes any hints on the correct proof would be so appreciated.

Your mistake is that from $\lim_{i\to\infty}\|v-u_i\|=dist(v,U)$ you have concluded that $u_i$ has a limit point. The limit above is in $\mathbb R$, but this does not imply that the original sequence $u_i$ has any limit points.
However, if $W$ is finite-dimensional, then the set $\{u_i:i\in\mathbb N\}\subseteq W$ is bounded and therefore has a limit point. (See e.g. https://math.stackexchange.com/a/144491/491874 for discussion about why it is.) That is the missing ingredient.
• I got that why my conclusion is incorrect but still I don't see why $[u_i]$ is bounded. Jan 16, 2018 at 20:23
• Because of $\lim_{i\to\infty}\|v-u_i\|=dist(v,U)$, if you pick any $\varepsilon\gt 0$, from some point onwards all the $u_i$ belong to the ball with the centre in $v$ and radius $dist(v,U)+\varepsilon$.