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Let $K$ be a compact metric space. Let $\{U_{i}\}$ be an open covering of $K$. Prove that there exists a number $j>0$ such that any ball of radius $j$ is contained in some $U_{i}$.

Here is my attempt:

Let $\{U_{i}\}$ be a covering and let $\{U_{j}\}$ be a finite subcover. Let $j$ be the minimal radius from the balls of the finite subcover. This is the number we are looking for

Is it ok?

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There is a nice short proof here. –  Brian M. Scott Sep 2 '12 at 6:42
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2 Answers

For each point $x\in X$ there is an $\epsilon(x)=:\epsilon>0$ and a $j\in J$ such that $U_\epsilon(x)\subset U_j$. The family $\bigl(U_{\epsilon/2}\bigr)_{x\in X}$ is an open covering of $X$; whence there is a finite collection of points $x_k\in X$ $\,(1\leq k\leq N)$ such that $x\subset\bigcup_{k=1}^N U_{\epsilon_k/2}(x_k)$. Let $\delta:=\min_{1\leq k\leq N}{\epsilon_k\over 2}>0$.

Assume now that $B\subset X$ is an open ball with center $p$ and radius $\delta$. There is a $k\in [N]$ such that $p\in U_{\epsilon_k/2}(x_k)$, and by the triangle inequality it follows that $B\subset U_{\epsilon_k}(x_k)\subset U_j$ for some $j\in J$.

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This doesn't work. Consider for example the unit interval $[0,1]$ with an open cover $[0,6/10),(5/10,1]$.

A proof that does work: Let $\{U_j\}$ be an open subcover and define $f_j(x)=\sup\limits_{\delta\geq 0}\{B_\delta(x)\subseteq U_j\}$, and note that this is continuous. Let $g(x)=\min\{f_j(x)\}$ and note that this is continuous and nonnegative, since $\{U_j\}$ is an open cover of $K$. Thus $\inf\limits_{x\in K}g(x)>0$ by compactness, and it follows that any open ball of radius at most $\inf\limits_{x\in K}g(x)$ is in some $U_j$.

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I see. Maybe then I shoud take the intersetion of Uj and the set the minimum over all intersections. I think then it should work. –  Karim Jonson Sep 2 '12 at 4:06
    
In fact,on a second thought,it doesn`t. –  Karim Jonson Sep 2 '12 at 4:30
    
@KarimJonson I added a proof to my answer, hope it makes sense. –  Alex Becker Sep 2 '12 at 4:37
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