Let $(X,\|\cdot \|)$ be an infinite dimensional normed vector space and $V \subseteq X$ a finite dimensional subspace, and $y \in X\backslash V$. I try to prove, that there is some $v \in V$ s.t.
$dist(y,V) = \inf_{v \in V} \| y – v\| > 0 $.
I tried the proof via contradiction:
Assume $\forall v \in V: \| y – v \| = 0$. Then $\forall \epsilon >0 \; \exists B(y, \epsilon) \; \text{s.t.} \; v \in B(y, \epsilon)$. Since $V$ is finite dimensional subspace, $V$ is closed, therefore $X\backslash V$ open, so we have $B(y, \epsilon) \subseteq X\backslash V$. Hence $v \in X\backslash V$, but this contradicts to $v \in V$.
Some feedback would be appreciated, whether this goes in the right direction or I have some major issues with my idea.
Best Answer
Let $d=d(x,V)$, there exists $x_n\in V$ such that $lim_nd(x_n,x)=d$. This implies that there exists $N$ such that $n\geq N$ implies that $d(x,x_n)<d+1$, we deduce that for $n\geq N$, $d(x_N,x_n)\leq d(x_N,x)+d(x,x_n)<2d+2$ and $(x_n)_{n\geq N}$ is bounded since $V$ is finite dimensional, there exists a subsequence $x_{n_q}$ of $x_n$ which converges towards $y$. $d(y,x)=d$.