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I recently stumbled upon this:

  • For any infinite countable subset $A\subseteq\mathbb R$ such that $\overline A=\mathbb R$, the complement $\mathbb R\setminus A$ is homeomorphic to the Baire space. (Or, equivalently, to the irrational numbers, i.e. $\mathbb R\setminus A\cong \mathbb R\setminus \mathbb Q$.)
  • If $A_1$, $A_2$ are both infinite countable dense subsets of $\mathbb R_2$, then $\mathbb R\setminus A_1\cong \mathbb R\setminus \mathbb A_2$, i.e., their complements are homeomorphic.

Here we take $\mathbb R\setminus A$ with the usual Euclidean topology.

Are there some proofs of these facts which are relatively straightforward? (Perhaps easier than the approaches described below.) Or even more generally, I'd be interested in various proofs and references for this result.

We could obtain this result from Alexandrov-Urysohn theorem. This theorem says that any nonempty Polish zero-dimensional space for which all compact subsets have empty interior is homeomorphic to Baire space $\mathcal N$. (See, for example, Theorem 7.7 in Classical Descriptive Set Theory by Kechris. This result is also mentioned in this answer: Polish space in which the interior of each compact set is empty.)

The space $B=\mathbb R\setminus A$:

  • Is Polish, since it is a $G_\delta$ subset of $\mathbb R$.
  • A base consisting of clopen sets can be obtained as $\{(a,b)\cap B; a,b\in A\}$.
  • We can use the fact that $A$ is dense to show that every compact set has empty interior.

Alternatively, if we look at the Arnold W. Miller's proof that irrationals are homeomorphic to $\mathcal N$, we might observe that the only facts about $\mathbb Q$ which is needed for this proof to go through is that $\mathbb Q$ is an infinite countable dense subset of the real line. This proof is given in this answer (which also includes some references): Baire space homeomorphic to irrationals.

I tried to find whether this has already been discussed on this site. I found only the following questions - which are tangentially related, but different.

Martin Sleziak
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  • There is a much stronger property, of course, also with a more complicated a proof: for each countable, dense subset of the reals there is a homeomorphism $f: \mathbb R \rightarrow \mathbb R$ which maps this set onto $\mathbb Q$. Hence its complement onto the irrationals. See here, 14 for a survey article of this and related features. – Ulli Jun 17 '24 at 06:54
  • @Ulli The property you mentioned is again a corollary of the approach in my answer: any order-isomorphism from a countable dense subset of $\mathbb{R}$ onto $\mathbb{Q}$ (which exists by Cantor’s isomorphism theorem) extends to an order-automorphism of $\mathbb{R}$, which is then necessarily a homeomorphism as the topology on $\mathbb{R}$ is the order topology. – David Gao Jun 17 '24 at 08:02

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A subset of $\mathbb{R}$ is dense in the topological sense iff it is order-dense. And any dense subset has its order topology coinciding with the inherited subspace topology, so an order isomorphism is automatically a homeomorphism. Also, any cocountable subset of $\mathbb{R}$ is automatically dense. So the results you mentioned are easy consequences of the lemma contained in my answer here, as well as the fact that all countable, dense linear orders without endpoints are order-isomorphic to each other.

David Gao
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