$A_\epsilon=[0,1]\bigcap\bigcup_{j=1}^\infty N_{1/2^j}(q_j)$. Show that $\partial A_\epsilon =[0,1]\setminus A_\epsilon$. 
Let $\{q_k\}$ be an enumeration of the rational numbers in $[0,1]$, and, for $0 < \epsilon<\frac{1}{2}$, let
  $$A_\epsilon=[0,1]\bigcap\bigcup_{j=1}^\infty \left(q_j-\frac{\epsilon}{2^j}, q_j+\frac{\epsilon}{2^j}\right).$$
  Show that $\partial A_\epsilon =[0,1]\setminus A_\epsilon$.

It looks to me that $A_\epsilon=[0,1]$. If we take the infinite union, $\bigcup_{j=1}^\infty \left(q_j-\frac{\epsilon}{2^j}, q_j+\frac{\epsilon}{2^j}\right)$, will we not get $[0,1]$?
How can I proceed?
 A: Let $x \notin A_\varepsilon$ and $r > 0$. Clearly $x \in B(x,r)$. On the other hand, we have that $$B(x, r)\cap \mathbb{Q} = \Big([0,1] \cap \langle x - r, x + r\rangle\Big) \cap \mathbb{Q} \ne \emptyset $$
because $\mathbb{Q}$ is dense in $\mathbb{R}$. In particular $B(x, r) \cap A_\varepsilon \ne \emptyset$ since $\mathbb{Q} \cap [0,1] \subseteq A_\varepsilon$.
Therefore $x \in \partial A_\varepsilon$ since an arbitrary ball around $x$ intersects both $A_\varepsilon$ and $A_\varepsilon^c$.
Now take $x \in A_\varepsilon$. Therefore, there exists $j \in \mathbb{N}$ such that $x \in \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle \cap [0,1]$ which is an open set in $[0,1]$ so there exists $r > 0$ such that $B(x,r) \subseteq \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle \cap [0,1]$. Therefore $B(x,r)$ does not intersect $A_\varepsilon^c$ so $x \notin \partial A_\varepsilon$.
We conclude $\partial A_\varepsilon = A_\varepsilon^c = [0,1] \setminus A_\varepsilon$.   

Indeed $A_\varepsilon \ne [0,1]$ for $\varepsilon < \frac12$ because if $\lambda$ denotes the Lebesgue measure, we have: $$\lambda\left(A_\varepsilon\right) \le \lambda\left(\bigcup_{j\in\mathbb{N}} \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle\right) \le \sum_{j=1}^\infty \lambda\left(\left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle\right) = \sum_{j=1}^\infty \frac{\varepsilon}{2^{j-1}} = 2\varepsilon < 1$$
and $\lambda([0,1]) = 1$.
