$D$ is compact, $U_{1},...,U_{n}$ are open, cover $D$ , then $\exists\delta>0$ s.t $\forall$ $x,y\in D $ with $|x-y|<\delta$, $x,y\in U_{j} $

Question

Let $D\in\mathbb{R}^{n}$ be a compact set. Suppose $U_{1},...,U_{n}$ are open subsets of $\mathbb{R}^{n}$ such that $D\subseteq\bigcup_{i=1}^{n}U_{i}$. Then there exists $\delta>0$ such that for all $x,y\in D$ with $|x-y|<\delta$, $x,y\in U_{j}$ for some $j$.

I am having trouble with this one. I tried proving it by contradiction, but I wasn't able to as well. Obviously the compactness plays a role. I was thinking that you could argue that if there never existed such a $\delta$, for all $p$ we have $D\nsubseteq \bigcup_{i=1}^{n}U_{i}-U_{p}$, and then this would contradict the compactness. But I wasn't able to.

Answer 1

For each $z\in D$, pick $i=i(z)$ with $z\in U_i$ and then $r=r(z)>0$ with $B_{r(z)}(z)\subseteq U_{i(z)}$. Then the $B_{r(z)/2}(z)$, $z\in D$, cover $D$. From a finite subcover thereof, let $\delta$ be the smallest radius used.

Then for $x,y$ with $|x-y|<\delta$, there is $z$ with $x\in B_{r(z)}$. Then $$|y-z|\le |x-y|+|y-z|< \delta+r(z)\le 2\delta.$$ As $B_{2\delta}(z)\subseteq U_{i(z)}$, the claim follows.

Answer 2

Hint

Assume that for any $n\in\mathbb{N}$ there exist $x_n,y_n\in D$ such that $|x_n-y_n|<1/n$ and $x_n, y_n\not\in U_i,$ for any $i.$ Now, since $D$ is compact there exists a subsequence $(x_{n_k})$ converging to $x\in D.$ Then it easy to show, using triangular inequality, that $(y_{n_k})$ also converges to $x.$ Now, $x\in U_i$ for some $i.$ Can you finish?