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Is it true that all countable subsets of $\mathbb{R}$ are homeomorphic (in their relative topology) to a subspace of $\mathbb{Q}$?

If this is true, then essentially the 'most complicated' countable set of reals we can think of (at least topologically) is $\mathbb{Q}$.

Lorenzo
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user754647
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    The definition of a sequence is precisely that of a function $\mathbb{N}\to X$, where $X$ is the space where the sequence takes values. Since there is a bijection between $\mathbb{N}$ and $\mathbb{Q}$, it is clear that we could index our sequence by $\mathbb{Q}$ instead of $\mathbb{N}$. I'm unsure what your question is asking, but maybe you could look into the concept of a net if you want to think about "more general sequences". See these notes: https://users.math.msu.edu/users/banelson/teaching/920/nets.pdf. –  Feb 13 '23 at 16:47
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    How do you put a topology on a sequence? – Vivaan Daga Feb 13 '23 at 16:48
  • @Peter I mean, it is always a numerable set, then bijective with a subset in $\mathbb{N}$. – user754647 Feb 13 '23 at 16:51
  • @Shinrin-Yoku I am thinking in the relative topology in $\mathbb{R}$. – user754647 Feb 13 '23 at 16:53
  • @OwenGary I am thinking in a sequence as the set of points in $\mathbb{R}$ in this case. – user754647 Feb 13 '23 at 16:55
  • @Lorenzo $\mathbb{N}$ has the discrete topology as a subset in $\mathbb{R}$. Hence, $\mathbb{Q}$ is not homeomorphic to $\mathbb{N}$. – user754647 Feb 13 '23 at 16:58
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    @user754647 I think that a proper formulation of the question would have been: "Is it true that every countable subset of $\mathbb{R}$ is homeomorphic to a subset of $\mathbb{Q}$"? A sequence and its range are different objects. – Lorenzo Feb 13 '23 at 17:05
  • @Lorenzo You are right! My initial question employed an abuse of notation. – user754647 Feb 13 '23 at 17:14
  • Since the subspace topology is an order topology, you can get an embedding as sketched here. – ChesterX Feb 13 '23 at 17:36
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    @user754647 the answer is affirmative and much more general, look here. – Lorenzo Feb 13 '23 at 17:38

1 Answers1

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Let $S$ be any countable subset of $\mathbb R$. Then $S \cup \mathbb Q$ is a countable dense subset of $\mathbb R$. I will define a homeomorphism $f$ of $\mathbb R$ to itself that takes $S \cup \mathbb Q$ onto $\mathbb Q$. Let $\{x_n\}_{n = 1}^\infty$ be an enumeration of $S \cup \mathbb Q$, and $\{y_n\}_{n =1}^\infty$ an enumeration of $\mathbb Q$. Define $f$ on $S \cup \mathbb Q$ inductively as follows.

Suppose $f(x_m)$ has been defined for all $m < n$. The numbers $x_m$ for $m < n$ partition $\mathbb R \backslash \{x_m: m < n\}$ into $n$ open intervals, and similarly $f(x_m)$ for $m < n$ partition $\mathbb R \backslash \{f(x_m): m < n\}$. If $x_n$ is in the $j$'th interval of $\mathbb R \backslash \{x_m: m < n\}$ (counting from left to right), let $f(x_n)$ be the first $y_k$ that is in the $j$'th interval of $\mathbb R \backslash \{f(x_m): m < n\}$.

By construction, $f$ is increasing, i.e. if $x_i < x_j$ then $f(x_i) < f(x_j)$. Moreover, it is onto: if $y_k$ hasn't yet been assigned as a value of $f$ when all $y_j$ for $j < k$ have been, say once $f(x_m)$ have been assigned for $m < n$, $y_k$ is in one of the intervals of $\mathbb R \backslash \{f(x_m): m < n\}$, and it will be assigned as $f(x_k)$ for the first $x_k$ in the corresponding interval of $\mathbb R \backslash \{x_m: m < n\}$. Then $f$ has a unique extension to a continuous increasing function from $\mathbb R$ onto itself.

Robert Israel
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