First of all I'd like to mention that it is a part of my home work so I'd like if you won't give the answer itself, but try to guide me into it. I've been losing my mind for the last couple of hours because I couldn't solve it on my own.
The question is as follows:
Assume that $X$ is a finite dimensional normed space, and $C$ is a closed nonempty subset (not necessarily a subspace) of $X$. Assume that there is $y\in C$ that $\|y\|=\inf\{\|x\| : x\in C\}$. Prove that for every $x\in X$ there is $c\in C$ that $d(x,c)=d(x,C)$.
I obviously tried to play a little with sequences $(y_n)$ and $(x_n)$ in $C$ that $$\lim_{n\to \infty} \|x_n\|=d(x,C)\quad \text{ and }\quad\lim_{n\to \infty}\|y_n\|=\|y\|,$$ but I had no luck.
I would love help. I truely lost my mind with this question, and I bet it is not even that hard at all... Thanks
First of all, note that if $X$ is finite-dimensional then the norm of $X$ is a closed map (this fails in infinite dimensions). Indeed, let $C$ be a closed set in $X$ and take $(w_n)_{n=1}^\infty$ in $C$ such that $\|w_n\|\to t$ as $n\to \infty$. Then the sequence $(w_n)_{n=1}^\infty$ is bounded, hence it contains a convergent subsequence $(w_{n_k})_{k=1}^\infty$ with limit $w$, say. Consequently, $t=\|w\|$.
Take $x\in X$ and choose a sequence $(u_n)_{n=1}^\infty$ in $C$ so that $\|x-u_n\| \to d(x, C)$ as $n\to \infty$. Choose now a convergent subsequence $(x-u_{n_k})_{k=1}^\infty$ and you are done. Consequently, $(u_{n_k})_{k=1}\infty$ converges too; let $u$ be its limit. As $C$ is closed, $u\in C$ and we are done. $\square$