I'm doing an undergrad project on the Banach-Tarski paradox and I'm not convinced by the proof I have come up with for this, everything to do with the Banach-Tarski Paradox is new maths to me and so I need to "dumb it down" a bit and really explain what I'm doing in each step. I've asked my supervisor if asking on stack is okay (as most sources have very brief proofs or less direct proofs using theorems I don't have space for in my report) and they said that it would be fine as long as I cite this if I use it. The statement I want to prove:
A set $X$ is $G$-paradoxical if and only if there exists disjoint subsets $A,B$ of $X$ such that $A \sim_G X$ and $B \sim_G X$.
In Stan Wagon's book The Banach-Tarski Paradox this is proven using the Banach– Schroder–Bernstein Theorem, but I do not have space to introduce and prove this in my report and so I'm trying to find a more direct proof. The proof I have so far uses the idea that we can rearrange the sets in a paradoxical decomposition in such a way that it gives us a partition and thus the paradoxical decomposition gives an equidecomposability relation, but I'm struggling to make a formal argument for the first implication.
My proof so far, the bold part being where I am struggling to generalize:
Suppose the set $X$ is $G$-paradoxical. Then, we have the paradoxical decomposition given by, \begin{equation} X = \bigcup\limits_{i=1}^{n} [ g_{i} \triangleright A_{i}] = \bigcup\limits_{j=1}^{m} [ h_{j} \triangleright B_{j}] \end{equation} Let $A = \bigcup \{A_i\}$, $B = \bigcup \{B_j\}$ and $X = \bigcup \{X_k\}$. Then, $A$ and $B$ are disjoint as $\{A_i\}\sqcup\{B_j\}$ are pairwise disjoint. Now, suppose $g_{i_1}, g_{i_2} \in \{g_i\}$ and $A_{i_1}, A_{i_2} \in \{A_i\}$ are such that \begin{equation*} A^{'} = g_{i_1} \triangleright A_{i_1} \cap g_{i_2} \triangleright A_{i_2} \neq \emptyset \text{, and } g_{i_1} \triangleright A_{i_1} \cup g_{i_2} \triangleright A_{i_2} = X_1 \cup X_2 \end{equation*} Then, we have \begin{gather} g_{i_1}^{-1} \triangleright A^{'} = A_{i_1} \cap (g_{i_1}^{-1} g_{i_2}) \triangleright A_{i_2} \\ \implies A_{i_1} \backslash \{g_{i_1}^{-1} \triangleright A^{'}\} = A_{i_1} \backslash \{A_{i_1} \cap (g_{i_1}^{-1} g_{i_2}) \triangleright A_{i_2}\}\\ \implies g_{i_1} \triangleright (A_{i_1} \backslash \{g_{i_1}^{-1} \triangleright A^{'}\}) = g_{i_1} \triangleright A_{i_1} \backslash \{g_{i_1} \triangleright A_{i_1} \cap g_{i_2} \triangleright A_{i_2}\}\\ \implies g_{i_1} \triangleright (A_{i_1} \backslash \{g_{i_1}^{-1} \triangleright A^{'}\}) = g_{i_1} \triangleright A_{i_1} \backslash A^{'} \end{gather} And so, we see that \begin{equation} \{g_{i_1} \triangleright A_{i_1} \backslash A^{'}\} \cap \{g_{i_2} \triangleright A_{i_2}\} = \emptyset \text{, and } \{g_{i_1} \triangleright A_{i_1} \backslash A^{'}\} \cup \{g_{i_2} \triangleright A_{i_2}\} = X_1 \cup X_2 \end{equation}
Thus, more generally, if we let $A_{i}^{'} = g_i \triangleright A_i \cap g_{i+1} \triangleright A_{i+1}$ and $A_{i}^{''} = \{A_i \backslash A_{i}^{'}\}$ it is possible to ensure that each $\{g_i \triangleright A_{i}^{''}\}$ are pairwise disjoint whilst maintaining $X = \bigcup\limits_{i=1}^{n} [g_i \triangleright A_{i}^{''}]$. Therefore, we can always construct the decomposition of $X$ such that the pieces form a partition of $X$. Hence, we can write $A \sim_G X$. The same argument is used to show $B \sim_G X$.
Conversely, suppose $A \sim_G X$ and $B \sim_G X$. Then we can partition the sets $A$ and $B$ as $A = A_1 \sqcup A_2 \sqcup ... \sqcup A_n$ and $B = B_1 \sqcup B_2 \sqcup ... \sqcup B_m$ such that for some $\{g_i\},\{h_i\} \subseteq G$ we have \begin{gather} g_i \triangleright A_i = X_i \hspace{0.2cm} \forall i \in \{1,2,...,n\} \text{ and,}\\ h_j \triangleright B_j = X_j \hspace{0.2cm} \forall j \in \{1,2,...,m\}. \end{gather} Where $X = X_1 \sqcup X_2 \sqcup ... \sqcup X_n$ and $X = X_1 \sqcup X_2 \sqcup ... \sqcup X_m$ are partitions of $X$. Thus we can write \begin{equation} X = \bigcup\limits_{i=1}^{n} [ g_{i} \triangleright A_{i}] = \bigcup\limits_{j=1}^{m} [ h_{j} \triangleright B_{j}]. \end{equation} Hence, $X$ is $G$-paradoxical.