Show that $\mathfrak{S}=\bigcup_{N=1}^{\infty}\mathfrak{Z}_N\cup\left\{\emptyset\right\}$ is a semi-ring 

Let $\Gamma$ be a finite set, $\Omega=\Gamma^{\mathbb{N}}=\left\{(x_1,x_2,\ldots):~\forall i\in\mathbb{N} x_i\in\Gamma\right\}$. For $a_1,\ldots,a_N\in\Gamma$ let
    $$
[a_1,\ldots,a_N]:=\left\{(x_1,x_2,\ldots)\in\Gamma^{\mathbb{N}}: i=1,\ldots,N x_i=a_i\right\}
$$
    be the $N$-cylinder which is determined by $a_1,\ldots,a_N$. Define
    $$
\mathfrak{Z}_N:=\left\{[a_1,\ldots,a_N]: a_1,\ldots,a_N\in\Gamma\right\}.
$$
    Show, that then
    $$
\mathfrak{S}:=\bigcup_{N=1}^{\infty}\mathfrak{Z}_N\cup\left\{\emptyset\right\}
$$
    is a semi-ring for $\Omega$.


Hello!
Three things are to show:
(1) $\emptyset\in\mathfrak{S}$
(2) $A,B\in\mathfrak{S}\implies A\cap B\in\mathfrak{S}$
(3) $A,B\in\mathfrak{S}$ and $A\subset B\implies~\exists A_1,\ldots,A_n\in\mathfrak{S}$ pairwise disjoint, so that $B\setminus A=A_1\cup\cdots\cup A_n$.
Proof. (1) is clear by definition of $\mathfrak{S}$.
(2) $A\in\mathfrak{S}$, i.e. $A=[a_1,\ldots,a_N]$ for a $N\in\mathbb{N}$ and $a_1,\ldots,a_N\in\Gamma$. $B\in\mathfrak{S}$, i.e. $B=[b_1,\ldots,b_M]$ for a $M\in\mathbb{N}$ and $b_1,\ldots,b_M\in\Gamma$. To my opinion then
$$
A\cap B=\begin{cases}A, & N\leq M\wedge a_i=b_i, i=1,\ldots,N\\B, & M\leq N\wedge b_i=a_i, i=1,\ldots,M\\\emptyset, & \text{otherwise}\end{cases}
$$
and $A,B,\emptyset\in\mathfrak{S}$.
(3) $A,B\in\mathfrak{S}, A\subset B$. If $A\subset B$, this means for $A=[a_1,\ldots a_N]$ and $B=[b_1,\ldots,b_M]$ that $N\leq M$ and $a_i=b_i, i=1,\ldots,N$. I am not sure, but to my opinion then $B\setminus A=\emptyset$. And so $B\setminus A$ can be writte as disjoint union of ONE set, namely the emptyset.

Would be great to know if my proof is ok.
Miro
 A: Your last point is not totally right, because if we have $A\subset B$, then $M\leq N$ (where the notation is the same as yours).
However, acording to this definition of semi ring, the third condition doesn't require that $A\subset B$ (I don't see right now if they are equivalent formulations, but I'll work in the more general outset). 
So, let $A, B \in \mathfrak{S}$ and let $A=[a_1,\ldots a_N]$ and $B=[b_1,\ldots, b_M]$. Assume without loss of generality, that $N\leq M$. If $a_i=b_i$ for every $1\leq i\leq N$, then $B\subset A$ and we have that $B\setminus A=\emptyset$
If $b_i\neq a_i$ for some $1\leq i\leq N$ then all the sequences in $B$ have as its $i$-the entry $b_i$ and all the sequences in $A$ have as its $i$-the entry $a_i$, so $A$ can't share any sequence with $B$ and we would have $B\setminus A=B$.
I don't see a flaw in your other points, so I think they are right.
A: On 3:
I preassume that $\Gamma$ has at least $2$ elements.
Then $\left[a_{1},\dots,a_{N}\right]\subseteq\left[b_{1},\dots,b_{M}\right]$
implies that $M\leq N$ and $a_{i}=b_{i}$ for every $i\in\left\{ 1,\dots,M\right\} $. 
So actually we have $\left[b_{1},\dots,b_{M}\right]=\left[a_{1},\dots,a_{M}\right]$.
If $\Delta:=\Gamma^{N-M}\setminus\left\{ \left[a_{M+1},\dots,a_{N}\right]\right\} $ then $\Delta$ is a finite set (since $\Gamma$ is) and we can write:
$\left[a_{1},\dots,a_{M}\right]\setminus\left[a_{1},\dots,a_{N}\right]=\bigcup_{\left[x_{M+1},\dots,x_{N}\right]\in\Delta}\left[a_{1},\dots,a_{M},x_{M+1},\dots,x_{N}\right]$
The RHS is a disjoint union of cylinder sets.

Another question brought me to this question.
