In my probability course, symmetric sets are Borel-measurable sets defined as follows:
$$ \begin{equation} \mathcal{S}=\{ S\in \mathcal{B}(\mathbb{R}^\mathbb{N}):p(S)=S \text{ for any finite permutation $p$}\} \end{equation}\tag{1}$$
where a finite permutation is a mapping $p:\mathbb{R}^\mathbb{N} \rightarrow \mathbb{R}^\mathbb{N}$ of the first $n$ elements of $x \in \mathbb{R}^\mathbb{N}$ such that:
$$ \begin{equation} p(x)=p(x_1,x_2,...)=(x_{k_1},...,x_{k_n},x_{n+1},...)\end{equation}\tag{2}$$
Now given that a symmetric set contains elements where the first $n$ terms are invariant to any permutation and $n$ may be as large as we want, I deduce that $\mathcal{S}$ is a sigma algebra in:
$$ \begin{equation} \Omega=\{\bar{x}: x \in \mathbb{R}\}\end{equation}\tag{3}$$
where $\bar{x}=(x,x,...x,...) \in \mathbb{R}^{\mathbb{N}}$.
On the other hand, given that the shift is a mapping $s:\mathbb{R}^\mathbb{N} \rightarrow \mathbb{R}^\mathbb{N}$ such that $\forall x \in \mathbb{R}^{\mathbb{N}}$:
$$ \begin{equation} s(x)=s(x_1,x_2,...)=(x_2,x_3,...)\end{equation}\tag{4}$$
shift-invariant sets are defined as follows:
$$ \begin{equation} \mathcal{I}=\{ I\in \mathcal{B}(\mathbb{R}^\mathbb{N}):s^{-1}I=I \} \end{equation}\tag{5}$$
Now, given that the shift operator may be iterated we have:
$$ \begin{equation} \lim_{n \to \infty} s^nI=I \end{equation}\tag{6}$$
and I think this last point is sufficient to conclude that $\mathcal{I}=\mathcal{S}$ but according to my professor this is not the case. Is there a non-trivial error at any step of my reasoning?
As pointed out by Kavi Rama Murthy, the following set is symmetric but not shift invariant:
$$\begin{equation} \{ (a_n) \in \mathbb{R}^{\mathbb{N}}: \sum a_n < 1 \} \end{equation} \tag{*}$$
Another example would be:
$$\begin{equation} \{ (a_n)\in \mathbb{R}^{\mathbb{N}}: \prod a_n < 1 \}=\{ (a_n) \in \mathbb{R}^{\mathbb{N}}: e^{ \sum \ln a_n} < 1 \} \end{equation} \tag{**}$$
so $(3)$ in my question is actually false and this is my original source of error. The set $\Omega=\{\bar{x}:x \in \mathbb{R}\}$ that I defined concerns the shift-invariant sets $\mathcal{I}=\sigma(\Omega)$. In fact, it is straightforward to show that $ \mathcal{I}\subset \mathcal{S}$.