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There is a problem from a list suggested practice problems that I am having issues with. It says:

Suppose that $X$ is a subspace of the real line $\mathbb{R}$ which is homeomorphic to the space of irrational numbers. Is the complement of $X$ in $\mathbb{R}$ necessarily countable?

Would anyone be willing to help me out with this one? Thank you so much!

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    The complement is not necessarily countable (as easily seen by Asaf's hint), but it *is* necessarily dense (because $X$ is a zero-dimensional subspace of a connected space).2012-09-02

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Hint: $\mathbb R$ is homeomorphic to $(0,1)$.

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    @Hagen: An easier way to achieve the same: start with the usual Cantor set $C$ and take its complement $X = \mathbb{R} \smallsetminus C$. Write $X$ as a countable disjoint union of intervals and remove the rationals from each. Finish up by noting that a countable disjoint union of the space of irrationals is still homeomorphic to the space of irrationals.2012-09-02