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Let $(U_i)_{i \in I}$ be an open cover of a metric space $(X,d)$, a number $\epsilon >0$ is called a Lebesgue number of $(U_i)_{i \in I}$ if for all $x \in X$ exist $j \in I$ such that $B(x,\epsilon) \subset U_j$. Show that every open cover of a compact set has a Lebesgue number.

I've came to the following proof, and wanted to check it was OK. (I'm not sure if I'm negating correctly in the "absurd" argument.)

Let's suppose that $X$ is compact and let $(U_i)_{i \in I}$ be an open cover of $X$. Let's suppose that $(U_i)_{i \in I}$ doesn't have a Lebesgue number, then $\forall \epsilon>0$ there exist $x_0 \in X$ such that for all $j \in I$, $B(x_0,\epsilon) \nsubseteq U_j$. Now since $X$ is compact, there exist a finite sub cover of $(U_i)_{i \in I}$, $\{U_{i_k}:k\in\{1,..,n\}\}$ such that $X=\bigcup_{k=1}^{n}U_{i_k}$. Then there exist $k_0 \in \{1,..,n\}$ such that $x_0 \in U_{i_{k_{0}}}$. Since $U_{i_{k_{0}}}$ is open, there exist $\epsilon_0 > 0$ such that $B(x_0,\epsilon_0) \subseteq U_{i_{k_{0}}}$ which is a contradiction.

What do you think?

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  • $\begingroup$ I don't think it's right. In your proof $\epsilon_0$ depends on $x_0$ while the assumption states 'for each $\epsilon$ there is a $x_\epsilon$ such that $B(x_\epsilon,\epsilon)$ does not fit in any $U_i$'. Now you have found an $\epsilon_0$ such that $B(x_0,\epsilon_0)$ in a certain $U_i$. But there could be a different $B(x,\epsilon_0)$ which does not fit in any $U_i$. $\endgroup$ – dietervdf Jun 6 '15 at 18:28
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Since $X$ is compact, there is a finite set $J\subset I$ such that $X=\bigcup_{j\in j} U_j$. Define $f:X\to\mathbb R$ by $$f(x)=\sup\{\delta>0, j\in J : B(x,\delta)\subset U_j\}.$$ Then $f$ is a continuous function defined on a compact metric space, so it attains a minimum value. Let $\varepsilon = \min_x f(x)$. That is the Lebesgue number.

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  • $\begingroup$ Why is this $f$ continous? $\endgroup$ – dietervdf Jun 6 '15 at 18:22

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