A Möbius transformation to prove that Cantor set is a homogeneous space The following Wikipedia's link
https://en.wikipedia.org/wiki/Cantor_set
affirms that "the Cantor set is a homogeneous space in the sense that for any two points $x$ and $y$ in the Cantor set $\mathcal {C}$, there exists a homeomorphism $ h:{\mathcal {C}} \mapsto {\mathcal {C}}$ with $h(x)=y$. These homeomorphisms can be expressed explicitly, as Möbius transformations."
Somebody knows this Möbius transformation? 
I know a demonstration of the homogenity of Cantor set without using Möbius transformations. 
 A: That line in the Wikipedia page is nonsense.  There are in fact only countably many Möbius transformations that map the Cantor set to itself (in fact, I suspect the only ones are $h(x)=x$ and $h(x)=1-x$).  To prove this, suppose $h(x)=\frac{ax-b}{cx-d}$ is a Möbius transformation which maps the Cantor set to itself.  Composing $h$ with $x\mapsto 1-x$ if necessary, we may assume $h$ is an orientation-preserving homeomorphism of $\mathbb{RP}^1$.  But this means that $h$ preserves the cyclic ordering of any triple of points in $\mathcal{C}$.  This means that if $x\in\mathcal{C}$ is the left endpoint of one of the triadic intervals used to construct $\mathcal{C}$, then $h(x)$ must also be such a left endpoint (since the left endpoints are exactly the points $x\in\mathcal{C}$ such that there exists $y\in\mathcal{C}$ such that there is no $z\in\mathcal{C}$ between $y$ and $x$).  This means there are only countably many possible values of $h(x)$.  Since a Möbius transformation is determined by its values on three points, we can determine $h$ by looking at its values on three left endpoints, and find there are only countably many possibilities for $h$.
