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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 $q+ \frac{\sqrt{2}}{n}\not\in\mathbb{Q}$.

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    A set $O$ is open if and only if $O=\mathrm{Int}(O)$. You can see that $\mathbb{Q}\neq\emptyset$, but $\mathrm{Int}(\mathbb{Q})=\emptyset$. To prove that $\mathbb{Q}$ isn't closed, let $(s_n)$ the sequence of the partial sums of the [decimal expansion](http://math.stackexchange.com/q/30062/8271) of $\sqrt{2}$. Then $s_n\in\mathbb{Q}$ for all $n$, but...2012-01-31

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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.

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    @Emir: yes, that was the argument: there are no interior points at all, because *no* non-empty open subset is a subset of the rationals or the irrationals.2012-01-31
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The ball inside Q doesn't fulfill the open set property B(x,r) subset of Q, because of dense property (It has Irrationals inside too)