Show that $\mathbb{Q}\subset\mathbb{R}$ with metric space $(\mathbb{R}, d)$ is neither open nor closed in $\mathbb{R}$.
Attempt: So I need to prove two parts.
(a) $\mathbb{Q}$ not open: Take $q\in\mathbb{Q}$. Note that $\frac{\sqrt{2}}{n}$, $n\in\mathbb{N}$ is arbitrarily small for arbitrarily large $n$, so $q\pm \frac{\sqrt{2}}{n}\in\mathbb{I}$. Thus for any $r>0$, $B_r(q)$ will contain irrational numbers so $B_r(q)\not\subseteq \mathbb{Q}$.
(b) $\mathbb{Q}$ not closed: We want to show $\mathbb{Q}$ does not contain all its accumulation points. $q+ \frac{\sqrt{2}}{n}\in\mathbb{I}$ for $n\rightarrow\infty$ is a limit point of $\mathbb{Q}$ but $q+ \frac{\sqrt{2}}{n}\not\in\mathbb{Q}$.
Best Answer
The basic fact to use is that every open interval in $\mathbb{R}$ contains both rational and irrational numbers (this is just reformulating that both sets are dense, essentially). This means no open set can be contained in just $\mathbb{Q}$ or $\mathbb{P}$ (the irrationals), so both sets have empty interior, and thus are far from being open.