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?
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.