$\bigcup_{x\in K}B'(x,r)=\left \{x\in X |d(x,K)\leq r\right \}$ for $K$ compact

Question

Let $(X,d)$ be a metric space and let $K$ be a compact subset of $X$

Let $r>0$ be given

Prove, $\bigcup_{x\in K}B'(x,r)=\left \{x\in X |d(x,K)\leq r\right \}$ where $d(x,K)=inf_{y\in K}d(x,y)$ (B' is closed ball)

Any hints ?

Answer 1

Hints:

$1).\ $ If $y\in \bigcup_{x\in K}B'(x,r),$ then $y\in B'(x,r)$ for some ball $B'$. Then, $d(x,y)\le r.$

$2).\ $ If $y\in \left \{x\in X |d(x,K)\leq r\right \},\ $ then fix this $y$ and note that the function $d(y,\cdot ):X\to \mathbb R$ is continuous, so it attains its minimum on the compact set $K$.