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There is a fact that countable subsets of the real line can exhibit very strange behaviors. For example, they can have no limit points, such as the set of integers $ \mathbb{Z} $. They can have a finite number of limit points, such as the set $ \{1/n\} $. They can even have uncountably many limit points, such as the set of rational numbers $ \mathbb{Q} $.

When we first encounter mathematical analysis, we may find ourselves wondering if we can classify the these countable subsets based on some equivalence relations "topologically", given the diversity of their properties.

One of the most straightforward equivalence relations that comes to mind is the following:

Let $ A \sim B $ if there exists a homeomorphism $f: \mathbb{R} \to \mathbb{R} $ such that $ f(A) = B $.

In the case of a finite number of limit points, it is relatively simple to determine whether two countable sets are equivalent under this relation. However, I am unsure how to approach more complex cases. For example, is there a case where two countably dense subsets (like $\mathbb Q$) are not equivalent under this relation?

My question: What are the known results or theories regarding the topological classification of countable subsets of $\mathbb R$ based on this or similar equivalence relations? This may be a strange question, but I think it’s still worth asking.

Zhang Yuhan
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    About your question typed in bold: yes, any two countable dense subsets of $\mathbb R$ are equivalent. This is the well-known property of "countable dense homogeneous" (CDH), see for instance here, and many more papers. – Ulli Sep 24 '24 at 10:52
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    Restricted to $\mathbb{R}$, you can use Cantor's isomorphism theorem to show that any two countable dense subsets are order-isomorphic, which implies homeomorphic since they use the order topology. Extending the homeomorphism to $\mathbb{R}$ is fairly straightforward from there since every element in the rest of $\mathbb{R}$ is determined by its place in the ordering. – MartianInvader Sep 25 '24 at 04:34
  • @MartianInvader: There is no isomorphism of the reals mapping $\Bbb Q$ to $\Bbb Q\cap(0,1)$. – Asaf Karagila Sep 26 '24 at 07:53
  • @AsafKaragila $\mathbb{Q} \cap (0,1)$ is not dense in $\mathbb{R}$. It's a little confusing because "dense" has two different meanings, but I'm assuming the question means topologically dense since it's trivial otherwise. Topologically dense in $\mathbb{R}$ implies dense as a linear order, but the converse doesn't hold. – MartianInvader Sep 26 '24 at 13:13

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Since all homeomorphisms $\mathbb R\to\mathbb R$ are bijective and continuous, and therefore order-preserving (or reverse the order), this boils down to involves classifying all countable linear orders (which can all be embedded into $\mathbb Q$). These are known to be very difficult to classify- see for example this post from MathOverflow for some partial results, as well as the explanations here. Note that classical order theory requires some understanding of set theory and ordinals to read, not out of necessity but because ordinals are intrinsic to the language used by set theorists studying ordered sets.

Edit: As pointed out by Giorgio, it gets even harder than this- I was mistakenly working under the assumption that the topology induced from the embedding of countable sets into $\mathbb R$ would always be the order topology- as it turns out, while in many examples this certainly is the case, there are also cases (e.g. $\{-1\}\cup\{\frac1n:n\in\mathbb N\}$) in which the embedding produces a strictly finer topology. So order-isomorphism is necessary, but not sufficient, for equivalence under homeomorphisms of $\mathbb R$.

Lavender
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    I believe it is even more difficult then classifying the linear orders since for example ${0}\cup{\frac{1}{n}:n\in\mathbb{Z}^+}$ and ${-1}\cup{\frac{1}{n}:n\in\mathbb{Z}^+}$ will be order isomorphic however they will not be equivalent. – Giorgio Genovesi Sep 25 '24 at 14:40
  • @GiorgioGenovesi I hadn't spotted that, thanks. – Lavender Sep 25 '24 at 15:48