I'm reading a book about Sidon sets and I'm stuck on the following proof. In order to facilitate the comprehension of my problems I will give the full proof and the context.
Let $G$ be a compact group and $\Gamma =\widehat{G}$ his dual group. A set $E\subset \Gamma \setminus \{1\}$ is called quasi-independant if for every choice for $m_j\in \{0,\pm 1\}$, $\prod _{j=1}^n\gamma _j^{m_j}=1$ (where $\gamma _j \in E$) implies $\gamma _j^{m_j}=1$ for every $j$. $E$ is a Sidon set if there is a constant $C$ such that for every finite set $F\subset E$ has a quasi-independant set $F'$ such that $|F|\leq C|F'|$.
Lemma Suppose that $E\subset \Gamma \setminus \{1\}$. Then there exists $\epsilon >0$ such that for every finite $F\subset E$, there is a set $Y=\{ y_1,\ldots y_N\}\subset G$ such that $N\geq 2^{\epsilon |F|}$ and $$\epsilon \leq \sup _{\gamma \in F}|\gamma (y_n)-\gamma (y_m)|, \forall n\neq m.$$
(The last inequality is called Pisier's $\epsilon $-net condition.)
We can assume that $F$ is quasi-independant since $A\subset F$ implies $$\sup _{\gamma \in A}|\gamma (x)-\gamma (y)|\leq \sup _{\gamma \in F}|\gamma (x)-\gamma (y)|.$$ Fix $0<\tau <1$ and pick $M_0$ such that $(1+\tau ^2)^{M_0}>8$ and assume $|F|\geq M_0$. Let $$X=\{ x\in G : \inf _{\gamma \in F}|\Re \gamma (x)|>\tau \} $$ and choose the largest integer $K\geq 2$ such $2^{K+1}<(1+\tau ^2)^{M}$ where $|F|=M$. For each non-trivial choice of $m=(m_1,\ldots ,m_K)\in \{0,1\}^K$, let $$X_m=\{ (x_1,\ldots ,x_K)\in G^K : \inf _{\gamma \in F}|\Re (\gamma (\prod _{k=1}^Kx_k^{m_k}))|>\tau \} .$$ If we let $s=\sum m_j$, then " it is easy to see that "
$$ X_m\subset \prod _{j :m_j\neq 0} \{ x_j : \inf _{\gamma \in F}|\Re (\gamma (x_j))|>\tau\}\times G^{K-s}. $$
I didn't understand this inclusion. If for every complex numbers $z_i$ we had $|\Re (\prod z_i)|\leq |\Re (z_i)|$, the inclusion woul'd be obvious, but this is not the case.
Assume this result is true. By using a previous lemma and the previous inclusion we can show that $m_{G^K}(\bigcup _{m\neq 0}X_m)<1$. This proves there exists $(x_1,\ldots ,x_K)\in G^K$ such that for all non-trivial $m=(m_1,\ldots ,m_K)\in \{0,1\}^K$ there exists $\gamma \in F$ with $|\Re (\gamma (\prod x_j^{m_j})|\leq \tau $.
And so $|\gamma (\prod x_j^{\pm m_j}-1|\geq 1- \tau$. I guess the idea is $$|\gamma (\prod x_j^{\pm m_j})-1| \geq |\Re \gamma (\prod x_j^{\pm m_j})-1 |\geq 1-|\Re \gamma (\prod x_j^{\pm m_j})|\geq 1- \tau $$ the last inequality were true if we had chosen $m_j \in \{0,\pm 1 \}$ and not $\{0, 1 \}$ ?!
Let $Y=\{ \prod _{j=1}^Kx_j ^{m_k} : m_k=0,1\}\setminus \{e\}$.(Shouldn't it be $m_k=0,\pm 1$ ?) The choice of $K$ ensures there is a constant $\epsilon _1$ wich not depend on $|F|$ so that $|Y|\geq 2^K-1\geq 2^{\epsilon _1|F|}$ (why are these products distincts ? $F$ is quasi-independant but not Y ?!). Then take any $\epsilon < 1-\tau$.
Any help will be greatly appreciated.