How to find an open cover $U$ for $(0,1)\subset\mathbb{R}$ with standard topology, such that $U$ does not have a Lebesgue number.

The question is: Find an open cover $U$ for $(0,1)\subset\mathbb{R}$ (with standard topology) such that: $\forall\epsilon>0$ $\exists x\in(0,1)$ such that $B(x,\epsilon)\not\subset u$ $\forall u\in U$. Or in other words there exists no Lebesgue number for $U$.

The only sets I can think of are closed sets. But the covering has to be open. I have no idea where to look for an answer. does one even exist? or is it a trick question?

For example $(0,\frac{1}{2}]\cup(\frac{1}{2},1)$ would work but left half is not open. any help or tips are much appreciated.

• covers of $[0,1]$ do have a Lebesgue number... – Henno Brandsma Apr 2 '17 at 20:41
• Try taking smaller and smaller balls around $\frac{1}{n}$ for all $n$, e.g. – Henno Brandsma Apr 2 '17 at 20:45
• Are you saying it is a trick question? – Peter Apr 2 '17 at 21:12
• Hint: consider the open cover $U = \big\{ (v,2v) \colon 0<v<\frac12 \big\}$. – Greg Martin Apr 2 '17 at 21:40
• Having the Lebesgue cover property implies being complete. – Henno Brandsma Apr 3 '17 at 3:20

For $n\in \mathbb N$ let $U_n=(2^{-n}, 3\cdot 2^{-n})\cap (0,1).$ If $2^{-n}<\epsilon/4$ and $x\in (0, 2^{-n})$ then for any $m$ such that $x\in U_m$ we have $m\geq n+1$ so the measure of $U_m$ is at most $2^{1-m}$ which is at most $2^{-n}$, but the measure of $B(x,\epsilon)\cap (0,1)$ is more than $2^{-n}.$
There is a theorem by Kasahara (for a proof, see Howes' book, "Modern Analysis and Topology" which says that $(X,d)$ has the Lebesgue property (every open cover of $X$ has a Lebsgue number) iff $X$ is a union of a compact and a discrete set. (in particular, such a space is complete)
So it's clear that $(0,1)$ must have such a cover, as it's not complete.