${(a_\lambda,b_\lambda)}_{\lambda \in \Lambda}$ is family set of open intervals ${(a_\lambda,b_\lambda)}_{\lambda \in \Lambda}$ is family set of open intervals for which $\bigcup^{}_{\lambda \in \Lambda}(a_\lambda,b_\lambda)\supset[0,1]$.
Prove that $\exists \epsilon$>0 $\forall x \in [0,1]$ $\exists \lambda_x \in \Lambda$ for which $(x-\epsilon, x+\epsilon)\subset(a_{\lambda_x},b_{\lambda_x}).$
I think that this problem is decided with assuming the opposite.
 A: So as to avoid making the problem trivial, I’ll give a sketch of the proof (at least how I did it), and leave filling in the details to you.
I use the following results:


*

*if $X$ is a compact set, and $\{O_\lambda\}_{\lambda\in\Lambda}$ is an open cover for $X,$ then there exists a finite $\Lambda_0\subseteq\Lambda$ such that $\{O_\lambda\}_{\lambda\in\Lambda_0}$ is an open cover for $X.$

*the max/min of finitely many continuous, real-valued functions is again continuous

*if $X$ is compact and $f:X\rightarrow\mathbb{R}$ is continuous, then there exist $x,y\in X$ such that $$f(x)=\min_{z\in X}f(z)\text{  and  }f(y)=\max_{z\in X}f(z).$$
We start by applying the first of the results above to reduce the open cover to a finite subcover $\Lambda_0$. Now we define for each $x\in[0,1]$ $$\Lambda_0(x):=\{\lambda\in\Lambda_0:a_\lambda<x<b_\lambda\},$$ and the function $$f(x):=\max_{\lambda\in\Lambda_0(x)}\max\{\varepsilon>0:(x-\varepsilon,x+\varepsilon)\subseteq(a_\lambda,b_\lambda)\}.$$ Since each $\Lambda_0(x)$ is finite, we observe that $f(x)>0$ for all $x\in[0,1].$ We want to prove that $f:[0,1]\rightarrow\mathbb{R}$ is continuous, and then applying the third of the results above, there is some point $x\in[0,1]$ such that $\min_{z\in[0,1]}f(z)=f(x)>0.$ This will complete the proof.
Now to show that $f$ is continuous, we first notice that we can rewrite $f$ as $$f(x)=\max_{\lambda\in\Lambda_0(x)}\min\{|x-a_\lambda|,|x-b_\lambda|\}.$$ Since $y\mapsto\max_{\lambda\in\Lambda_0(x)}\min\{|y-a_\lambda|,|y-b_\lambda|\}$ is continuous using the second result above twice, we are nearly there, but we need to show that if $|x-y|$ is small, then (1) $\Lambda_0(y)\supseteq\Lambda_0(x)$, and (2) if $\Lambda_0(y)\neq\Lambda_0(x),$ that for $\lambda’\in\Lambda_0(y)\setminus\Lambda_0(x),$ $\min\{|y-a_{\lambda’}|,|y-b_{\lambda’}|\}<f(y),$ so that only the $\lambda\in\Lambda_0(x)$ contribute to the outer maximization in $f(y)$ (using the second formula). I found the quantity $2\delta:=\min_{\lambda\in\Lambda_0(x)}\min\{|x-a_\lambda|,|x-b_\lambda|\}$ to be helpful for this part.
Note that we do have to consider (2) above, since we may have some $a_\lambda$ or $b_\lambda\in[0,1]$, and if $x=a_{\lambda’}$ (say), then for any positive $\eta,$ there will be $y$ with $|y-x|<\eta,$ but $\Lambda_0(y)\neq\Lambda_0(x).$ As we would expect, however, the $\lambda\in\Lambda_0(x)$ which achieves the maximum in the second formula for $f(x)$ will contain a larger ball around $y$ than $(a_{\lambda’},b_{\lambda’})$ will, when $|x-y|$ is small enough.
A: This is a simple application of Heine Borel Theorem. Using Heine Borel theorem we can see that a finite number of intervals $(a_\lambda, b_\lambda) $ cover $[0,1]$. Consider the set of all these chosen finite number of values $a_\lambda, b_\lambda $ which lie in $[0,1]$. These points together with $0,1$ form a partition of $[0,1]$. Let $0\in(a_{\lambda_{0}},b_{\lambda_{0}})$ and $1\in(a_{\lambda_{1}},b_{\lambda_{1}})$ and further let $\delta$ be the length of the smallest sub-interval created by the partition of $[0,1]$ described earlier. If we take a positive number $$\epsilon <\min(\delta/2,-a_{\lambda_{0}}/2,(b_{\lambda_{1}}-1)/2)$$ then $\epsilon $ is the desired number (I hope you can see why this is the case). 
A: If $r$ is a Lebesgue number for the cover, then every set of diameter $r$ is contained in some interval $(a_{\lambda},b_{\lambda})$ of the cover. So I think you can take $\epsilon$ to be $r/2.$
