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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). –  tomasz Sep 2 '12 at 18:26
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1 Answer

up vote 8 down vote accepted

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

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Much simpler than the counterexample I was constructing to give an $X\subset(0,1)$ homeomorphic to the irrational numbers! I think I won't bother with mine, now.... XD –  Cameron Buie Sep 2 '12 at 16:37
Great! What if we add the condition that $X$ is dense in $\mathbb R$? –  Hagen von Eitzen Sep 2 '12 at 16:49
@Hagen: every compact subset of $\mathbb{N^N}$ has empty interior, hence it's nowhere dense. Remove a Cantor set from the irrationals and the resulting space will still be dense in $\mathbb{R}$. –  t.b. Sep 2 '12 at 17:12
@Hagen: Every Polish space which is $\sigma$-compact, and every compact subset has an empty interior is homeomorphic to the irrationals. It is not hard to verify that by removing a Cantor set from the irrationals these properties are unchanged. –  Asaf Karagila Sep 2 '12 at 17:23
@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. –  t.b. Sep 2 '12 at 17:44
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