It is true that $\overline{(0,\epsilon)\cup(\mathbb Q\cap(\epsilon,1))}=[0,1]$ for $\epsilon\in(0,1)$? Let $\epsilon\in(0,1)$ and $E=(0,\epsilon)\cup(\mathbb Q\cap(\epsilon,1))$. It is true that $\overline{E}=[0,1]$?
I know that $\overline{E}\subset[0,1]$, but how to show that $\overline{E}\supset[0,1]$?
I've tried to show that if I have an open interval $(a,b)$ then $\overline{\mathbb Q\cap(a,b)}=[a,b]$. By the way, is this true?
 A: The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
\overline{E} = \overline{(0,\varepsilon) \cup (\mathbb{Q}\cap(\varepsilon,1))}=\overline{(0,\varepsilon)}\cup\overline{\mathbb{Q}\cap(\varepsilon,1)} = [0,\varepsilon] \cup \overline{\mathbb{Q}\cap(\varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $\overline{\mathbb{Q}\cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $\overline{\mathbb{Q}\cap(a,b)} \subseteq \overline{\mathbb{Q}}\cap\overline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x \in [a,b]$, any neighborhood of $x$ intersects the set $\mathbb{Q}\cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $\mathbb{Q}\cap(a,b)$.
If $x \in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $\mathbb{Q}$ then implies the existence of a rational number $q \in (x-r,x+r)$, and by construction this rational number is in $\mathbb{Q}\cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) \subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q \in (x,x+r)$, therefore in $\mathbb{Q}\cap(a,b)$, so $(x-r,x+r)$ intersects $\mathbb{Q}\cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
