Prove that if $\mathit{K \subset X}$ is complete and $x \notin \mathit {K}$, then $\exists k \in \mathit {K}$ such that $d(x,k)=d(x,\mathit{K})$

Question

I'm very confused. First I tried to show that $d_\mathit{K} (x) = d(x, \mathit{K})$ is continuous and if $\mathit{K}$ is bounded then exists $k$, but $\mathit{K}$ complete doesn't imply $\mathit{K}$ bounded.

Also, If there is a sequence that converges to $k$, I'm not sure this helps.

I was thinking in a counterexample, where $\mathit{K}=\Bbb{R}$ and take a point of $\Bbb{C}$ (i.e. $x = i$), but i don't know if it's the best way.

Could anyone help me?

Answer 1

A concrete counterexample: Let $X=l^\infty(\mathbb N)$ and$$K=\{(a_i)_j|i\in\mathbb N^+ \text{ and } \forall j\in\mathbb N^+, a_{ij}=(1+\frac{1}{j})\delta_{ij}\}$$

$K$ is closed: For any two points $x,y\in K$, $d(x,y)=\|x-y\|> 1$. Therefore any Cauchy sequence will eventually stabilize.

And it's easy to see that $d(0, K)=1$, but no $k\in K$ satisfies $d(0,k) = 1$.

However, you cannot find such an example with $\mathbb R$ (isometrically embedded into any $X$), because the closed ball $\bar B(x, d(x,K)+1)$ has nontrivial intersection with $K$, and therefore the intersection is a nonempty compact subset in which if $d(x, k_i)\rightarrow d(x, K)$, then we can find a convergent subsequence $k_{i_j}$ such that $k_{i_j}\xrightarrow{j\rightarrow\infty} k\in K$ and $d(x,K)=d(x,k)$ follows.