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Is all of R the only open set containing Q?

False, take any irrational $p$ in $R$. Then $Q\subset R\setminus\{p\}$ and $R\setminus\{p\}$ is open. Of course we can take any subset A of $R\setminus Q$ that is closed in R and take $R\setminus A$.

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
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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
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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

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up vote 3 down vote accepted

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 $$\mathbb{Q}_{\text{open cover}} = \bigcup_{k=0}^{\infty} \left(q_k - \dfrac{\epsilon}{2^{k+2}},q_k + \dfrac{\epsilon}{2^{k+2}} \right)$$

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