+\begin{proposition}
+The function $d$ defined in Eq.~\ref{nouveau d} is a metric on $\mathcal{X}$.
+\end{proposition}
+
+\begin{proof}
+ $d_e$ is the Hamming distance. We will prove that $d_s$ is a distance
+too, thus $d$ will be a distance as sum of two distances.
+ \begin{itemize}
+\item Obviously, $d_s(S,\check{S})\geqslant 0$, and if $S=\check{S}$, then
+$d_s(S,\check{S})=0$. Conversely, if $d_s(S,\check{S})=0$, then
+$\forall k \in \mathds{N}, |S^k\Delta {S}^k|=0$, and so $\forall k, S^k=\check{S}^k$.
+ \item $d_s$ is symmetric
+($d_s(S,\check{S})=d_s(\check{S},S)$) due to the commutative property
+of the symmetric difference.
+\item Finally, $|S \Delta S''| = |(S \Delta \varnothing) \Delta S''|= |S \Delta (S'\Delta S') \Delta S''|= |(S \Delta S') \Delta (S' \Delta S'')|\leqslant |S \Delta S'| + |S' \Delta S''|$,
+and so for all subsets $S,S',$ and $S''$ of $\llbracket 1, \mathsf{N} \rrbracket$,
+we have $d_s(S,S'') \leqslant d_e(S,S')+d_s(S',S'')$, and the triangle
+inequality is obtained.
+ \end{itemize}
+\end{proof}
+
+
+Before being able to study the topological behavior of the general
+chaotic iterations, we must firstly establish that:
+
+\begin{proposition}
+ For all $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, the function $G_f$ is continuous on
+$\left( \mathcal{X},d\right)$.
+\end{proposition}
+
+
+\begin{proof}
+We use the sequential continuity.
+Let $(S^n,E^n)_{n\in \mathds{N}}$ be a sequence of the phase space $%
+\mathcal{X}$, which converges to $(S,E)$. We will prove that $\left(
+G_{f}(S^n,E^n)\right) _{n\in \mathds{N}}$ converges to $\left(
+G_{f}(S,E)\right) $. Let us remark that for all $n$, $S^n$ is a strategy,
+thus, we consider a sequence of strategies (\emph{i.e.}, a sequence of
+sequences).\newline
+As $d((S^n,E^n);(S,E))$ converges to 0, each distance $d_{e}(E^n,E)$ and $d_{s}(S^n,S)$ converges
+to 0. But $d_{e}(E^n,E)$ is an integer, so $\exists n_{0}\in \mathds{N},$ $%
+d_{e}(E^n,E)=0$ for any $n\geqslant n_{0}$.\newline
+In other words, there exists a threshold $n_{0}\in \mathds{N}$ after which no
+cell will change its state:
+$\exists n_{0}\in \mathds{N},n\geqslant n_{0}\Rightarrow E^n = E.$
+
+In addition, $d_{s}(S^n,S)\longrightarrow 0,$ so $\exists n_{1}\in %
+\mathds{N},d_{s}(S^n,S)<10^{-1}$ for all indexes greater than or equal to $%
+n_{1}$. This means that for $n\geqslant n_{1}$, all the $S^n$ have the same
+first term, which is $S^0$: $\forall n\geqslant n_{1},S_0^n=S_0.$
+
+Thus, after the $max(n_{0},n_{1})^{th}$ term, states of $E^n$ and $E$ are
+identical and strategies $S^n$ and $S$ start with the same first term.\newline
+Consequently, states of $G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are equal,
+so, after the $max(n_0, n_1)^{th}$ term, the distance $d$ between these two points is strictly less than 1.\newline
+\noindent We now prove that the distance between $\left(
+G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is convergent to
+0. Let $\varepsilon >0$. \medskip
+\begin{itemize}
+\item If $\varepsilon \geqslant 1$, we see that distance
+between $\left( G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is
+strictly less than 1 after the $max(n_{0},n_{1})^{th}$ term (same state).
+\medskip
+\item If $\varepsilon <1$, then $\exists k\in \mathds{N},10^{-k}\geqslant
+\varepsilon > 10^{-(k+1)}$. But $d_{s}(S^n,S)$ converges to 0, so
+\begin{equation*}
+\exists n_{2}\in \mathds{N},\forall n\geqslant
+n_{2},d_{s}(S^n,S)<10^{-(k+2)},
+\end{equation*}%
+thus after $n_{2}$, the $k+2$ first terms of $S^n$ and $S$ are equal.
+\end{itemize}
+\noindent As a consequence, the $k+1$ first entries of the strategies of $%
+G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are the same ($G_{f}$ is a shift of strategies) and due to the definition of $d_{s}$, the floating part of
+the distance between $(S^n,E^n)$ and $(S,E)$ is strictly less than $%
+10^{-(k+1)}\leqslant \varepsilon $.\bigskip \newline
+In conclusion,
+$$
+\forall \varepsilon >0,\exists N_{0}=max(n_{0},n_{1},n_{2})\in \mathds{N}%
+,\forall n\geqslant N_{0},
+ d\left( G_{f}(S^n,E^n);G_{f}(S,E)\right)
+\leqslant \varepsilon .
+$$
+$G_{f}$ is consequently continuous.
+\end{proof}
+
+
+
