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Let $(X,\tau)$ a compact metric space and $\{ U_i : i \in I \}$ an open cover of $X$. Show that there is $r>0$ such that for all $a \in X$ there is an $i \in I$ such that $B_{r}(x) \subseteq U_{i}$.

My attempt:

By definition of compactness, $X$ is covered by some finite subset of $\{ U_{i} : i \in I \}$. Let $U_{1}, \ldots , U_{n}$ be such a finite subcover of $X$.

Choose any $x\in X$. Suppose $x$ lies in $U_1$. There is some number $r_{1}(x)$ such that for any $r < r_{1}(x)$, the ball $B_{r}(x)$ lies in $U_{1}$. For any $x \in X$, we may associate to $x$ the $n$ numbers $r_{1}(x), \ldots, r_{n}(x)$, noting that at least one of these is non-negative. Let $r(x)$ be the least non-negative member of $\{ r_{1}(x), \ldots , r_{n}(x) \}$.

Below, I prove that $r: X \to \mathbb{R}$ is a continuous function. As $r$ is continuous, $r(X)$ is a continuous subset of the real numbers. Therefore $r(X)$ is a finite union of finite closed intervals in $\mathbb{R}$. In particular, $r(X)$ contains a least element which we denote by $r_{0}$. Note that as $r(x)$ is greater than zero for all $x \in X$, we know that $r_{0} > 0$. For any $p and for any $x\in X$, we know that $B_{p}(x)$ lies in at least one $U_{i}$.


Is this correct? Can you give me another alternative solution?

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    This is very close to the result known as [Lebesgue's covering lemma](http://en.wikipedia.org/wiki/Lebesgue%27s_covering_lemma) or [Lebesgue's number lemma](http://www.proofwiki.org/wiki/Lebesgue%27s_Number_Lemma). There are a few questions on MSE related to this, e.g., [Proof of the Lebesgue number lemma](http://math.stackexchange.com/questions/82240/proof-of-the-lebesgue-number-lemma), [Uses of Lebesgue's covering lemma](http://math.stackexchange.com/questions/65721/uses-of-lebesgues-covering-lemma) or [Explanations of Lebesgue number lemma](http://math.stackexchange.com/questions/105337/).2012-06-26

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