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I have been stuck in the following problem for a few weeks and I have not been able to prove it.

Fix $D_{1}$ and $D_{2}$ disjoint countable dense subsets of $2^{\omega}$ such that $\mathcal{D}=D_{1}\cup D_{2}$ is an independent family. Why are there at least $\mathfrak{c}$ homeomorphisms $f_{\eta}:G_{\eta}\longrightarrow G_{\eta}$ such that $f[D_{1}]=D_{2}$, where $G_{\eta}\supseteq\mathcal{D}$ is a $G_{\delta}$ subset of $2^{\omega}$?

Fernando Mauricio Rivera Vega
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In the following, I assume that the sets $G_\eta$ are not fixed. At least, given the subindex $\eta$, it seems to be the case.

Since $D_1$ and $D_2$ are dense in $\mathcal{D}$, which in turn is dense in $2^\omega$, it follows that they are perfect, countable metric spaces and thus homeomorphic to $\mathbb{Q}$. On the other hand, any homeomorphism between $D_1$ and $D_2$ extends to a homeomorphism between $G_\delta$ subsets, by Lavrentiev's theorem. Hence, it is enough to construct $\mathfrak{c}$ homeomorphisms from $\mathbb{Q}$ onto itself.

Let $z\in \left\{\frac{1}{3},\frac{2}{3}\right\}^\omega$, and define $$ H_z(x) :=\begin{cases} x & x<0 \\ n+2z_n(x-n) & x\in \left[n,n+\frac{1}{2}\right]\cap\mathbb{Q} \\ n+1+2(1-z_n)(x-n-1) & x\in \left[n+\frac{1}{2},1\right]\cap\mathbb{Q}, \end{cases} $$ where $n$ ranges over the nonnegative integers (you can see this idea put to use in this paper on homeomorphisms of the real line).

Then $\{H_z\}_z$ is a family of $\mathfrak{c}$ homeomorphisms of $\mathbb{Q}$.

Pedro Sánchez Terraf
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  • Thank you very much for your response, it helped me a lot. The only thing I have a problem with now is that, Lavrentiev's theorem extends to me homeomorphism between two sets $G_\delta $ not necessarily equal and just that I need. – Fernando Mauricio Rivera Vega Aug 10 '19 at 18:39
  • There is a version of Lavrentiev's theorem for homeomorphisms $\mathcal{D}\to\mathcal{D}$, and that one gives an extension to a map $G\to G$ (Exercise 3.10 in Kechris). On the other hand, I'm noticing that in my solution, $\mathcal{D}$ might not be included in any of the two $G_\delta$ sets. The problem is more delicate. – Pedro Sánchez Terraf Aug 10 '19 at 20:18
  • A minor remedy: Repeat the construction in the answer but starting from $\mathcal{D}$ and $D_2$ instead of $D_1$ and $D_2$. At least you'll have that $\mathcal{D}$ is included in the domain of the homeomorphisms. – Pedro Sánchez Terraf Aug 11 '19 at 11:44