I am working on the following task:
Show that for every $\epsilon>0$ there is an closed set $A\subseteq\mathbb{R}$ with $\lambda^1(\mathbb{R}\setminus A)\leq\epsilon$, that does not contain a nonempty open interval.
This is what I have so far: Take $\mathbb{Q}=(q_n|n\in\mathbb{N})$ and define $A:=\mathbb{R}\setminus\bigcup_{n\in\mathbb{N}}(q_n-r_n,q_n+r_n)$ with appropiate $r_n>0$ for $n\in\mathbb{N}$.
I am not sure how to contiune from here. Can anybody help me with this?
This is the complete proof, thanks to everyone who helped with this:
Look at $\mathbb{Q}=(q_n|n\in\mathbb{N})$. Let's define $A$ as follows $A:=\mathbb{R}\setminus\bigcup_{n\in\mathbb{N}}(q_n-r_n,q_n+r_n)$ with $r_n>0$ $\forall n\in\mathbb{N}$. Let's now find a $r_n$ that helps to fulfill the requirements, for this we want: $$\sum_{n\in\mathbb{N}}r_n\leq\frac{\epsilon}{2}$$ For this we easily find $r_n=\frac{\epsilon}{2^{n+1}}$ $\forall n\in\mathbb{N}$. It now follows: $$\lambda^1(\mathbb{R}\setminus A)=\lambda^1(\bigcup_{n\in\mathbb{N}}(q_n-r_n,q_n+r_n))$$ Since $\lambda^1$ is $\sigma$-subadditive it follows $$\lambda^1(\mathbb{R}\setminus A)\leq\sum_{n\in\mathbb{N}}\lambda^1((q_n-r_n,q_n+r_n))=\sum_{n\in\mathbb{N}}(q_n+r_n-q_n+r_n)=\sum_{n\in\mathbb{N}}2r_n=\epsilon$$ $A$ is closed as the complement of an open set. $A$ does not contain any nonempty interval since $A\cap\mathbb{Q}=\emptyset$, $A$ is disjoint from $\mathbb{Q}$. With this we have found a closed set $A\subseteq\mathbb{R}$ with $\lambda^1(\mathbb{R}\setminus A)\leq\epsilon$, that does not contain a nonempty open interval, this conlcudes the proof.