Given $E = \{x \in \mathbb{Q}: a < x < b\}$, prove that $[a,b] \subset \partial E$ I am currently studying for my real analysis midterms and as such, I am doing the older midterms. I have currently reached the question asked in the title, that is:
Given $E = \{x \in \mathbb{Q}: a < x < b\}$, prove that $[a,b] \subset \partial E$
Now, I have written down a proof, but I am not completely sure that it is valid and so I wanted to get an opinion here. The proof goes as follow:
First, we show that the interior of $\mathbb{Q}$ is empty. To do that, see that $\mathring{\mathbb{Q}} = \{x \in \mathbb{Q}: \exists \delta \in \mathbb{R}, \delta > 0, (x-\delta, x+\delta) \subseteq \mathbb{Q}\}$. However, there exists no such $\delta$ since $\mathbb{R}\setminus\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, $\mathring{\mathbb{Q}} = \emptyset$.
Secondly, it is given that $\overline{\mathbb{Q}} = \mathbb{R}$, and so $\partial \mathbb{Q} = \mathbb{R} \setminus \emptyset = \mathbb{R}$.
Finally, since $E \subset \mathbb{Q}$, we find by definition that $[a,b] \subset \partial E$.
And so, my question is: can I confidently say this or are the first and second steps not strong enough to ensure the last result?
 A: Pretty good.  I don't really like your final step though. I feel it could be clearer. 
Alternatively,  why not just set out to prove it directly? 
Given $x\in [a,b]$, we need $x\in\partial E$.  By definition,  $x\in \partial E$ if $\forall\epsilon\gt0$ we have $(x-\epsilon,x+\epsilon)\cap E\setminus \{x\}\neq\emptyset$.
This is true, since, firstly, (as you say) $\bar{\Bbb Q}=\Bbb R$.  That is,  $\Bbb Q$ is dense in $\Bbb R$.
In particular,  $E=\Bbb Q\cap (a,b)$ is dense in $[a,b]$.  That takes care of the irrationals in the interval. 
Put this together with the fact that  there is a sequence of rationals converging to any rational. That is,  we need to know any rational in $[a,b]$ is a limit point of $E$.
For, $x\in \partial E$ iff $x$ is a limit point of $E$ (this is true in general).
A: I think you have the right idea, but if your reasoning isn't convincing yourself, then you need to add a little bit more. I would try appealing straight to the definitions. Recall that $\partial E = \overline{E} \setminus E^\circ$, so you need to show that $[a, b] \subset \overline{E}$ and $[a, b] \cap E^\circ = \emptyset$.
As per your first step, the interior of $\Bbb{Q}$ is empty. If $c \in E^\circ$, then there would exist some $r > 0$ such that $B(c; r) \subset E \subset \Bbb{Q}$, which would put $c \in \Bbb{Q}^\circ = \emptyset$. Thus, $[a, b] \cap E^\circ = [a, b] \cap \emptyset = \emptyset$.
Your second step really does need some fleshing out. You will, of course, use the density of rationals, but density doesn't survive the process of taking subsets necessarily. Just knowing $\overline{\Bbb{Q}} = \Bbb{R}$ doesn't tell us too much about the closure of a subset of $\Bbb{Q}$.
Suppose $x \in (a, b)$. Then, consider the open sets $(x - \delta / n, x + \delta / n)$, where $\delta = \min \{ x - a, b - x\}$. Then these open sets are contained in $(a, b)$. By the density of $\Bbb{Q}$ in $\Bbb{R}$, there exists some
$$x_n \in \Bbb{Q} \cap (x - \delta / n, x + \delta / n) \subset \Bbb{Q} \cap (a, b) = E.$$
But then $|x_n - x| < \delta / n$, hence $x_n \to x$. That is, $x \in \overline{E}$.
It only remains to show $a, b \in \overline{E}$. We can show $a \in \overline{E}$ by considering open sets of the form $\left(a, a + \frac{b - a}{n}\right)$, and similarly $b \in \overline{E}$ using sets of the form $\left(b - \frac{b - a}{n}, b\right)$. After you establish this, you're done.
