# Is all of $\mathbb{R}$ the only open set containing $\mathbb{Q}$?

Is all of $\mathbb{R}$ the only open set containing $\mathbb{Q}$?

False, take any irrational $p$ in $\mathbb{R}$. Then $\mathbb{Q} \subset \mathbb{R} \setminus \{p\}$ and $\mathbb{R} \setminus \{p\}$ is open.

Of course we can take any subset $A$ of $\mathbb{R} \setminus \mathbb{Q}$ that is closed in $\mathbb{R}$ and take $\mathbb{R} \setminus A$.

Is what I got right? or did I forget something?

What fact or theorem can I use?

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What is R{p}?... – DonAntonio Nov 7 '12 at 4:31
Your reasoning is valid. – Austin Mohr Nov 7 '12 at 4:34
Maybe you meant to write $R - \{p\}$ instead of $R\{p\}$. If so I agree with your example. – coffeemath Nov 7 '12 at 4:34
Maximiliano wrote R\{p}. Markdown eats backslashes. – MJD Nov 7 '12 at 4:36
Yes. What you have written is correct. In fact, you can do much better in the sense that you can cover $\mathbb{Q}$ with open set with arbitrarily small length $\epsilon$, for instance $$\bigcup_{k=0}^{\infty} \left(q_k - \dfrac{\epsilon}{2^{k+2}},q_k + \dfrac{\epsilon}{2^{k+2}}\right)$$ – user17762 Nov 7 '12 at 4:58

Yes. What you have written is correct. In fact, you can do much better in the sense that you can cover $\mathbb{Q}$ with open set with arbitrarily small length $\epsilon$, for instance $$\mathbb{Q}_{\text{open cover}} = \bigcup_{k=0}^{\infty} \left(q_k - \dfrac{\epsilon}{2^{k+2}},q_k + \dfrac{\epsilon}{2^{k+2}} \right)$$