A question on limit points I have a query regarding a question from Stephen Abbott's Understanding analysis.
Question: Is $\mathbb{Q}$ an open or closed set, or neither? Given a point is $\mathbb{Q}$ is there an $\epsilon$-neighborhood contained in $\mathbb{Q}$?
The solutions suggest $\mathbb{Q}$ is neither open nor closed - fine. 
It also says that there is no $\epsilon$-neighborhood contained in $\mathbb{Q}$
I am not quite sure I understand the meaning of that last line.
For instance, isn't an $\epsilon$-neighbourhood centred at $0$ with $\epsilon = \frac{1}{10}$ contained in $\mathbb{Q}$ ?
 A: No. This neighbourhood contains the point $\frac{1}{10\sqrt{2}}$ which is not a rational.
A: An $\epsilon$ neighborhood around $0$ is the open interval $(-\epsilon,\epsilon)$. All intervals contain both rational and irrational numbers.
A: First of all, open or closed in what ? Because it's both open and closed in itself. ;-) This is general, any topological space $X$ is open and closed in itself, for open sets are stable by intersection, the empty intersection (which is therefore open) gives the empty set, which is open, so that its complement $X$ is closed in $X$, and the empty union gives the whole set $X$, which is open in $X$.
I guess you meant in $\mathbf{R}$. Then it is not closed because $\overline{\mathbf{Q}} = \mathbf{R} \not= \mathbf{Q}$ as $\sqrt{2}\not\in\mathbf{Q}$. And it is not open because then $0$ would be in its interior, so that you could find a small open ball $B(0,\varepsilon)$ contained in $\mathbf{Q}$, but this small ball will necessarily meet irrational numbers (for $\frac{\pi}{2^n}$ for $n\in\mathbf{N}$ big enough would be in it) leading to a contradiction.
