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Theorem. (Sierpiński) Any countable metric space without isolated points is homeomorphic to $\mathbb{Q}$.

Let $\mathbb{N} = \{1, 2, ...\}$, $\mathbb{Q}_- = \{r\in \mathbb{Q}: r<0\}$.

If $X$ has a finite number of isolated points then we should be able to embed it in $\mathbb{Q}_-\cup \mathbb{N}$ by placing the isolated points in $\mathbb{N}$, and using Sierpiński's theorem to create a homeomorphism of limit points of $X$ and $\mathbb{Q}_-$.

Similarly if $X$ has a finite number of limit points, then it should be homeomorphic to a finite disjoint union consisting of copies of $\{0\}\cup\{1/n : n\in\mathbb{N}\}$. But then we can easily embed each copy of $\{0\}\cup\{1/n : n\in\mathbb{N}\}$ in $[2m, 2m+1]\cap \mathbb{Q}$ for $m\in\mathbb{N}$.

What happens in the general case?

bof
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Jakobian
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1 Answers1

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I use "countable" to mean "finite or countably infinite".

Theorem. (Sierpiński) Any nonempty countable metric space without isolated points is homeomorphic to $\mathbb{Q}$.

Corollary. Any countable metric space is homeomorphic to a closed subspace of $\mathbb Q$.

Proof. If $X$ is a nonempty countable metric space, then $X$ is homeomorphic to a closed subspace of $X\times\mathbb Q$, and $X\times\mathbb Q$ is homeomorphic to $\mathbb Q$ by Sierpiński's theorem.

bof
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