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Question: Find a open cover for $\mathbb{R}$ that has no Lebesgue number.

My doubt: Well I am still strugling with this question. I recall that for a compact subset of metric space, every open cover has a lebesgue number.

Does anyone know how to do this?

Math
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BigBang
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1 Answers1

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For $n\in \mathbb{Z}$, take $A_n = (n,n+1)$ and $B_n := (n-1/n, n+1/n)$ and consider the open cover $\mathcal{U} := \{A_n :n \in \mathbb{Z}\} \cup \{B_n : n \in \mathbb{Z}\}$.

For any $\delta > 0, \exists N \in \mathbb{N}$ such that $2/N < \delta$, so consider $$ C := (N+1-1/N, N+1+1/N) $$ Then $\text{diam}(C) < \delta$, but $C$ is not contained in any single member of $\mathcal{U}$.

Prahlad Vaidyanathan
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    @BigBang: This answer is incorrect: $\mathscr{U}$ does not cover $0$; $B_0$ is undefined, and the endpoints of $B_n$ need to be interchanged when $n<0$. It can be salvaged by letting $$B_n=\left(n-\frac1{|n|+1},n+\frac1{|n|+1}\right);.$$ – Brian M. Scott Nov 28 '16 at 16:40