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Let $X$ be a compact metric space, and $\{U_{\lambda}:\lambda\in\Lambda\}$ an open cover of $X$. Let $\delta$ be its Lebesgue number. I want to show that the sets $F_\lambda=\{x\in U_\lambda:d(x,\partial U_\lambda)\leq \delta\}$ also cover $X$.

As a start, $x\in F_\lambda$ means that $x\in B(y,\delta) \cap U_{\lambda}$ for $y \in \partial U_{\lambda}$. I do not see why $F_\lambda$ cover $X$.

Thank you for any help.

:: I just noticed that as in the proof of the Lebesgue Number lemma, $d(x,\partial U_\lambda):=\inf_{y \in \partial U_\lambda}d(x,y)$.

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    It would make more sense to me if $F_\lambda$ were defined as $\{x\in U_\lambda:d(x,\partial U_\lambda)\geq \delta / 2 \}$. Then for every $x$, $B(x, \delta/2)$ is contained in some $U_\lambda$ and $x \in F_\lambda$.2012-05-14

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What if $X=[0,1]$ with the usual topology, and the open cover consists of the sets $U_0=[0,9/16)$ and $U_1=(1/2,1]\}$? We can take $\delta=1/16$, and then $F_0=[1/16,1/2]$ and $F_1=[9/16,15/16]$, which don't cover $X$.

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    If $\delta$ is a Lebesgue number for the open cover $U_{\lambda}$, then for all $x$ in $X$ $B(x,\delta) \subseteq U_{\lambda}$ for some single $\lambda$. $(-1/16,1/16)$ is not contained in $U_0$ and (15/16,17/16) is not contained in $U_1$,am I wrong?2012-05-04
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    @YellowMonkey: The space $X$ doesn't include any negative numbers or numbers greater than $1$, so those intervals are irrelevant. Their intersections with $X$ are $[0,1/16)$ and $(15/16,1]$, which are contained in $U_0$ and $U_1$, respectively.2012-05-04
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    Yes, of course the subspace topology. I am convinced by the counterexample. But do you have any idea how the question can be reformulated so that something remarkable is said? I should have said earlier but thank you very much for your response, Brian.2012-05-04
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    @YellowMonkey: I gave it a bit of thought, and I don't at the moment see any interesting way to salvage it, I'm afraid.2012-05-04