Some questions about Banach Tarski proof Banach-Tarski proof as been the topic of a video by the well-known Youtube channel VSauce but there were some parts that I didn't understand. So I went reading for the proof on Wikipedia, and I didn't understand the same parts too. It's about the countable infinity of "poles".

Here is a quick overview of the proof :


The infinite group with two genrators $\{A, B\}$ is called the free group $F$
Each element $X$ of $F$ can be bijectively mapped to a string $s(X)$ made of $a, b, a^{-1}, b^{-1}$ with no $aa^{-1}, a^{-1}a, bb^{-1}, b^{-1}b$ substring
If we note $\{x*\}$ the strings starting with $x$, and $e$ the empty string, then :

$F = s^{-1}(\{a*\} \cup \{b*\} \cup \{a^{-1}*\} \cup \{b^{-1}*\} \cup \{e\})$

$F = A^{-1}s^{-1}(\{a*\}) \cup s^{-1}(\{a^{-1}*\})$

$F = B^{-1}s^{-1}(\{b*\}) \cup s^{-1}(\{b^{-1}*\})$
One can chose two rotations $R$ and $T$ such that the action of their generated group $H$ on $\mathbb{S}_2$ is isomorphic to $F$.
If we note $M=\mathbb{S}_2/H$, then the equalities above rewrites as :

$\mathbb{S}_2 = s^{-1}(\{r*\} \cup \{r^{-1}*\} \cup \{t*\} \cup \{t^{-1}*\} \cup \{e\}) \times M$

$\mathbb{S}_2 = (R^{-1}s^{-1}(\{r*\}) \cup s^{-1}(\{r^{-1}*\})) \times M$

$\mathbb{S}_2 = (T^{-1}s^{-1}(\{t*\}) \cup s^{-1}(\{t^{-1}*\})) \times M$


What I don't understand how the poles are a problem. First the rotations of $H$ will have a countable infinity of poles $P$. Those cannot be elements of $H$ since there would then be a finite subgroup of $H$ and that's impossible given that $H$ is isomorph to $F$. So these poles are in a different equivalence class i.e. no fixed point string isn't it ? Then in the video, Michael states that there is a countable infinity of poles. But isn't the set of poles on the sphere $P \times M$ uncountable ? What are those troubles Wikipedia is talking about ?


  In Step 3, we partitioned the sphere into orbits of our group $H$. To streamline the proof, we omitted the discussion of points that are fixed by some rotation; since the paradoxical decomposition of $F$ relies on shifting certain subsets, the fact that some points are fixed might cause some trouble.
The poles are the points of $S^2$ that are left fixed by at least one nontrivial element of $H$. Since $H$ is countable and each element of $H$ has exactly two fixed points, there are countably many poles. 

In the presence of fixed points, the fact "every point in $S^2$ can be reached in exactly one way by applying the proper rotation from $H$" that is used in step 3 of the proof as given in Wikipedia is wrong.

Therefore, one needs to get rid of these poles - which happens by the same trick we can use to let a single point disappear as in $\mathbb N\to\mathbb N$, $n\mapsto n+1$.
The poles are the points of $S^2$ that are left fixed by at least one nontrivial element of $H$. Since $H$ is countable and each element of $H$ has exactly two fixed points, there are countably many poles. 

In the presence of fixed points, the fact "every point in $S^2$ can be reached in exactly one way by applying the proper rotation from $H$" that is used in step 3 of the proof as given in Wikipedia is wrong.

Therefore, one needs to get rid of these poles - which happens by the same trick we can use to let a single point disappear as in $\mathbb N\to\mathbb N$, $n\mapsto n+1$.