Let $n\geq 1$ be an integer and suppose $S\subseteq {\mathbb R}^n$ is countable and dense. Do we have $S \cong {\mathbb Q}^n$ where both sets carry the topology inherited from the Euclidean topology on ${\mathbb R}^n$?
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asked Sep 14, 2017 at 5:44
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4$\begingroup$ Related cool fact: If $n$ is allowed to be an infinite cardinal, then all countable dense subsets of $\mathbb R^n$ are homeomorphic if and only if $n < \mathfrak p$. (The question has already been answered, but I thought you might like to know.) $\endgroup$– Will BrianCommented Sep 14, 2017 at 17:43
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2$\begingroup$ And of corse everyone but me knows what $\mathfrak p$ is. $\endgroup$– Gerald EdgarCommented Sep 14, 2017 at 20:34
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2$\begingroup$ @GeraldEdgar: The Stone-Cech remainder of the naturals, $\mathbb N^*$, has the interesting property that every nonempty $G_\delta$ set has nonempty interior. What about intersections of larger cardinality? The cardinal number $\mathfrak p$ is, by definition, the smallest cardinality of a collection of open subsets of $\mathbb N^*$ that has a nonempty intersection with empty interior. That's a formal definition. An informal definition is that $\mathfrak p$ is the cardinal where certain diagonalization-type constructions stop working. To me, that's what makes it a really interesting number. $\endgroup$– Will BrianCommented Sep 15, 2017 at 8:36
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$\begingroup$ Amazing, thanks @WillBrian! And I guess we can make this result to hold even for all $n < {\frak t}$! :) $\endgroup$– Dominic van der ZypenCommented Sep 18, 2017 at 9:54
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2 Answers
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According to https://arxiv.org/abs/1210.1008 Example 2(c)... yes, they are all homeomorphic to $\mathbb Q$!
answered Sep 14, 2017 at 6:11
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12$\begingroup$ Oh, and the reason is Sierpinski's theorem stating that every countable, metrizable space without isolated points is homeomorphic to $\mathbb Q$. $\endgroup$ Commented Sep 14, 2017 at 6:22
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1$\begingroup$ Great - thanks for this concise answer, and it's a nice paper you pointed to! $\endgroup$ Commented Sep 14, 2017 at 6:43
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1$\begingroup$ Long ago, my friend Juliusz Jabłecki, then a high school student, wrote an article (which is apparently unavailable; I only found this), where he proved that any two countable subsets of a manifold are connected via a diffeomorphism. This preprint is cited, for example, in an article by Maciej Pietroń, which deals with Hilbert's cube. $\endgroup$ Commented Sep 14, 2017 at 9:40
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$\begingroup$ @MateuszKwaśnicki You mean dense countable subsets? $\endgroup$ Commented Sep 14, 2017 at 10:01
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$\begingroup$ @მამუკაჯიბლაძე: Yes, this is what I meant, thanks! $\endgroup$ Commented Sep 14, 2017 at 10:07
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You can learn a bit more about countable dense subsets of separable metric spaces by searching for "countable dense homogeneous space".
A separable metric space $X$ is Countable Dense Homogeneous (CDH) if given any two countable dense subsets $D$ and $E$ of $X$ there is a homeomorphism $f : X \rightarrow X$ such that $f(D) = E$.
The concept was introduced by R. Bennett in Countable dense homogeneous spaces Fund. Math., 74 (1972), pp. 189-194
Theorem 3 in the paper implies that locally euclidean spaces are CDH. So not only any countable dense subset of ${\bf R}^n$ is homeomorphic to ${\bf Q}^n$, but the homeomorphism can be taken as the restriction of a global homeomorphism of ${\bf R}^n$.
