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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!

MJD
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Maria
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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 02 '12 at 18:26

1 Answers1

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

Asaf Karagila
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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 02 '12 at 16:37
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    Great! What if we add the condition that $X$ is dense in $\mathbb R$? – Hagen von Eitzen Sep 02 '12 at 16:49
  • @Cameron: Were you thinking about Cantor set minus endpoints? – Asaf Karagila Sep 02 '12 at 17:06
  • @Hagen: That is an interesting question. My hunch is that the answer is no, I'll think of an argument and write another comment. – Asaf Karagila Sep 02 '12 at 17:08
  • Precisely, Asaf. – Cameron Buie Sep 02 '12 at 17:08
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    @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 02 '12 at 17:12
  • @t.b.: It may be dense in $\mathbb R$, but I don't see immediately that it's still homeomorphic to $\mathbb R\setminus \mathbb Q$. – Hagen von Eitzen Sep 02 '12 at 17:21
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    @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 02 '12 at 17:23
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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. – t.b. Sep 02 '12 at 17:44