Let $A$ be an open set in $\mathbb{R}$ with usual topology and $A \cup\mathbb Q=\mathbb{R}$. Does it imply $A=\mathbb{R}$ ?
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13$\mathbb{R} \setminus {0}$. – Martin May 17 '13 at 15:52
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No, I can remove any set of rationals that contains all its limit points.
So I could remove $\mathbb Z$ for example or $\{0\}\cup\{\frac 1n : n\in\mathbb N\}$ for example, or many more complex ones.
Tim
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3To illustrate the possible complexity: Tim's example is essentially the case $\alpha=\omega$ of the following: For any countable ordinal $\alpha$ there is a closed set of rationals whose order type is precisely $\alpha+1$. Its complement is open. – Andrés E. Caicedo May 17 '13 at 16:18