I have an exercise that wants me to construct a concrete example of a homeomorphism $$ \varphi:\mathbb{Q}\to\mathbb{Q}\times\{0,1\}\subset \mathbb{R}^2. $$ I have no idea where even to begin, I already have trouble finding a bijection, let alone a homeomorphism. Looking up some similar questions led me to the Sierpinski theorem, which states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. However, we haven't had that in our lecture, so I doubt that I'm allowed to use it.
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4Hint: if you take the set of rationals less than $\sqrt{2}$, it is homeomorphic to all of $\mathbb Q$. – Cheerful Parsnip Sep 28 '23 at 15:38
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Does this answer your question? Is every countable dense subset of $\mathbb R$ ambiently homeomorphic to $\mathbb Q$ Or this? Let $(P,<),(Q,\prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(P,<),(Q,\prec)$ are order-isomorphic – Anne Bauval Sep 28 '23 at 15:54
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@AnneBauval sadly not, as the exercise expects me to provide a concrete homeomorphism, so just saying that it exists by a theorem is not sufficient. Also, I dont know if $\mathbb{Q}\times{0,1}$ can be ordered, but maybe it can – Niels Slotboom Sep 28 '23 at 16:44
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1@NielsSlotboom For something explicit, take the lexicographic ordering. Abstractly, any set can be (well) ordered – FShrike Sep 28 '23 at 16:53
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2Hint: Using the induced topology from $\mathbb{R}^2$, we see that $\mathbb{Q}\times{0,1}$ is disconnected. Is $\mathbb{Q}$ disconnected? (yes) Can you write $\mathbb{Q}$ as a union of two open sets? Then, connect this idea to @CheerfulParsnip 's approach. – Michael Burr Sep 28 '23 at 16:54
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Related: $\Bbb Q\simeq\Bbb Q\setminus{0}.$ – Anne Bauval Sep 28 '23 at 19:06
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We will define a homeomorphism $h:\mathbb Q\times \{0,1\} \rightarrow \mathbb Q $ by defining maps on intervals. Define $$ f_n: (\pi+n, \pi+n+1) \times \{0\} \rightarrow (\pi+2n,\pi+2n+1)$$ by $(t,0) \mapsto t+n $ and $$ g_n: (\pi+n, \pi+n+1) \times \{1\} \rightarrow (\pi+2n+1,\pi+2n+2)$$ by $(t,1) \mapsto t+n+1$ for all $n \in \mathbb Z$. Clearly $f_n,g_n$ are homeomorphisms. Restricting all such $f_n,g_n$ on $\mathbb Q\times \{0,1\}$ we will obtain the desired homeomorphism.
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You say restricting, but you mean defining piecewise using $f_n, g_n$. – Jakobian Mar 06 '25 at 10:26
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