I have a rather simple question by I cannot find an answer anywhere.
Let $K$ be a compact subset of $\mathbb R^n$ and let $(O_i)_{i=1,..,n}$ be $n$ open balls that cover $K$. Can we reduce the radii $r_i$'s of $O_i$ a little bit such that the new open balls $(O_i')_{i=1,...,n} $ with radius $r'_i < r_i$, concentric with the original balls, and still cover $K$?
Any hints would be highly appreciated! Thank you!
On
Thank you @TheSilverDoe for the link, I write my answer here.
Let $m_i$ be the center of $O_i$. Let $x \in K $, then $x\in O_i$ for some $i \in \{1,...,n\}$, thus there is $r'_i< r_i$ such that $x \in O'_i$, concentric with $O_i$ with smaller radius.
Therefore, all the balls of the form $B(m_i,r'_i),$ for $ i \in \{1,...,n\}, r_i' \in (0,r_i)$, cover $K$. We extract a finite covering from this and discard redundant balls in the following way: if center $m_j$ appears several times then keep only the ball with biggest radius (which is smaller than $r_i$!).
Of course, if some center does not appear, then add the ball with that center and radius lower than the original one. The union of these balls (with smaller radius) now still covers $K$.
Yes. Let $A_{i,k}$ be the ball centered at the center of $O_i$, but with radius $k$ times that of $O_i$. Let $B_k=\bigcup_{i}A_{i,k}$. Let $X\subseteq[0,+\infty)$ be the set of $k$'s such that $K\setminus B_k\neq\emptyset$. By convention, let $0\in X$ since it is not clear what $B_0$ is. Let $Y=[0,+\infty)\setminus X$ be the set of $k$'s such that $K\subseteq B_k$. It suffices to prove that $Y$ is open. Equivalently, let's show that $X$ is closed. Obviously, $X$ and $Y$ are both intervals. We just want to show that they are not semi-open. So suppose that $X$ is semi open, of the form $[0,t)$ for some $t>0$. Then the sequence $B_{t-\frac{1}{n}}$ is a non-finite cover of $K$ since $\bigcup_{n\in\mathbb{N}}B_{t-\frac{1}{n}}=B_t$, and it is a linearly ordered (under subset relation) family so $K$ cannot be compact, a contradiction.