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I am having trouble figuring out the following question (3.10 in Kechris, Classical Descriptive Set Theory): If $X$ is completely metrizable, and $A\subseteq X$ with $f:A\to A$ a homeomorphism, then there is a $G_\delta$ set $G\subseteq X$ containing $A$ and an extension $h:G\to G$ of $f$ which is a homeomorphism of $G$.

Lavrentiev's Theorem gets us $G_\delta$ sets $G'$ and $H'$ containing $A$, and a homeomophism $g:G'\to H'$ extending $f$, but it doesn't seem to me that the proof of that result can be easily adapted to make $G'=H'$. The key fact in all of this is the set $\bar{A}\cap\{x:\mathrm{osc}_f(x)=0\}$, for any continuous $f:A\to X$, is a $G_\delta$ set in $X$ on which we can continuously extend $f$.

Iian Smythe
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  • Can't you simply work with $(f,f^{-1})\colon A \times A \to A \times A$ to make the situation symmetric modulo switching coordinates? – t.b. Jun 01 '12 at 23:23
  • @t.b. I don't see what you mean by "modulo switching coordinates". Could you explain? – Iian Smythe Jun 02 '12 at 01:17
  • I was a bit quick, sorry. The idea was that in the proof of Lavrentiev's theorem you intersect two graphs to get the mutually inverse homeomorphisms. I mistakenly assumed that you automatically get a symmetric situation when applying this to the mutually inverse homomorphisms $(f,f^{-1})$ and $(f^{-1},f)$ from $A\times A \to A \times A$. Unfortunately, this isn't quite the case and it seems that you need to perform a Schroeder-Bernstein type argument as given by Brian below to make this idea work. – t.b. Jun 02 '12 at 10:15

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By Lavrentiev’s theorem $f$ can be extended to a homeomorphism $f_0:G_0\to H_0$, where $G_0$ and $H_0$ are $G_\delta$-sets in $X$ containing $A$. Let $G_1=G_0\cap H_0$, and let $H_1=f_0[G_1]$; clearly $G_1$ is a $G_\delta$ in $X$. Moreover, $G_1$ is a $G_\delta$ in $G_0$, and $f_0$ is a homeomorphism, so $H_1$ is a $G_\delta$ in $H_0$ and therefore in $X$. Finally, $f_1\triangleq f_0\upharpoonright G_1:G_1\to H_1$ is a homeomorphism extending $f$. Now let $H_2=G_1\cap H_1$, $G_2=f_1^{-1}[H_2]$, and $f_2=f_1\upharpoonright G_2$; $G_2$ and $H_2$ are $G_\delta$-sets containing $A$, and $f_2$ is a homeomorphism between them extending $f$.

Continue in this fashion. If $n\in\omega$ is even, $G_{n+1}=G_n\cap H_n$ and $H_{n+1}=f_n[G_{n+1}]$, while if $n$ is odd, $H_{n+1}=G_n\cap H_n$ and $G_{n+1}=f_n^{-1}[H_{n+1}]$, and $f_{n+1}=f_n\upharpoonright G_{n+1}$ in both cases. The sets $G_n$ and $H_n$ are $G_\delta$-sets in $X$, and each $f_n$ is a homeomorphism from $G_n$ onto $H_n$ extending $f$. Note that $$H_0\supseteq G_1\supseteq H_2\supseteq G_3\supseteq H_4\supseteq\dots\;.$$

Now let $$G=\bigcap_{n\in\omega}G_n=\bigcap_{n\in\omega}H_n\qquad\text{ and }\qquad\bar f=\bigcap_{n\in\omega}f_n=f_0\upharpoonright G\;;$$ clearly $G$ is a $G_\delta$ containing $A$, and $\bar f$ is a homeomorphism of $G$ onto $$\bar f[G]=\bigcap_{n\in\omega}f_n[G_n]=\bigcap_{n\in\omega}H_n=G\;.$$

Brian M. Scott
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    I thought that some kind of dovetailing like this might be involved. Thanks! – Iian Smythe Jun 03 '12 at 03:04
  • How did you come up with such a technique? It is not easy for a novice. – Michael May 20 '22 at 16:35
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    @Michael: Agreed, it’s not easy for a novice, but this kind of back-and-forth technique is actually fairly common, so with experience it’s quite natural. One of the first such proofs that a novice is likely to see is this one, which is the heart of the proof that every regular, Lindelöf space is normal. The details are of course completely different, but the overall structure is similar. – Brian M. Scott May 21 '22 at 03:30