K compact subset. Closest point to y in K.

Question

If $ K $ is a compact subset of $\mathbb R^d$ and $y$ is a point of $\mathbb R^d$ which is not in $K$, then there is a closest point to $y$ in $K$. That is, there is an $x_0$ $\in$ $K$ s.t. $||x_0-y|| \le ||x-y||$ $\forall x\in K $ .

My attempt at figuring out solution. since $y \notin K$ this means that $y \notin$ $\bigcup O_\alpha$ and that $y$ is not contained in any finite subcover. I was thinking of using the complement of the union of the open cover and find a ball that contains $y$ and intersects the compact space $K$ at a point $x_0$.

I'm not sure if I'm missing something or if I should consider more? Hints and advice would be greatly appreciated.

Answer 1

Don't use the definition of compact. Use the fact that the function $f\colon K\longrightarrow\mathbb R$ defined by $f(k)=\|k-y\|$ must have a minimum, since it is continuous and $K$ is compact.

Answer 2

Hint: Prove that the function $x\mapsto \|x-y\|$ is continuous on compact $K$, obviously bounded from below (by $0$) and so it must reach a minimum.

Actually we have not even used the fact $y\not\in K$: the statement is valid for $y\in K$ too, except that it is trivial for such $y$: namely, you can take $x_0=y$.