Prove that $\mathcal{A}$ is an algebra Let $X=\{1,\ldots,m\}$ and
\begin{equation}
r:X\rightarrow [0,1] \\ i\mapsto r_i^s
\end{equation}
such that $\sum_{i=1}^\infty r_i^s=1$ (with $s\in \mathbb{R}$). Let $C\subset X^k=\{(i_1,\ldots,i_k):1\leq i_j\leq m\}$.
Define
\begin{equation}
I_{C}=\{(x_1,x_2,\ldots): (x_1,x_2,\dots,x_k)\in C\}.
\end{equation}
For $k=0$, define $I_{\emptyset}=X^k$. Prove that $\mathcal{A}=\{I_C: k\in \mathbb{N}, C\subset X^k\}$ is an algebra.
I wanted to understand how I can show that it is closed for completion, that is, given $I_m\in \mathcal{A}$ then how can I prove that $\mathcal{A}\setminus I_m\in \mathcal{A}$?
And besides, how can I show that given $I_m$ and $I_k$, with $k\neq m$, then $I_m\cup I_k\in \mathcal{A}$?
 A: $\def\N{\mathbb{N}}$
$\def\A{\mathcal{A}}$
The sets $I_C$ are a subset of $X^{\N}$. You thus want to prove

*

*$X^{\N} \in \A$ ;

*Given $C \subset I_k$, $X^{\N} \setminus I_C \in \A$ ;

*Given $C_1$ and $C_2$ subsets of $X^{k_1}$ and $X^{k_2}$, $I_{C_1} \cup I_{C_2}  \in \A$.

Let's do it !

*

*For $C = X \subset X^1$, we have $I_C = X^{\N}$, therefore $X^{\N} \in \A$.

*Let $k \in \N$ and let $C$ be a subset of $X^k$. We must show that $X^{\N} \setminus I_C \in \A$, so we must find a set $C'$ such that $I_{C'} = X^{\N} \setminus I_C$.

As $I_C$ is the set of all sequence whose first elements are given by a finite sequence from $C$, its complement is the set of all sequence whose first elements are not by a finite sequence from $C$. Therefore, One very natural candidate for $C'$ is to take $C' = X^{k} \setminus C$. Let's show that it works. Lets $x=(x_n)_{n \in \N} \in X^{\N}$. Then we have the equivalence :
$$\begin{array}{lll}
x \in I_{C'} &\Longleftrightarrow& (x_1,x_2,\dots,x_k) \in C' \\
&\Longleftrightarrow& (x_1,x_2,\dots,x_k) \notin C \\
&\Longleftrightarrow& x \notin I_{C} \\
&\Longleftrightarrow& x \in X^{\N} \setminus I_{C} \\
\end{array}$$
Therefore $I_{C'} = X^{\N} \setminus I_C$, and this step is over.


*I strongly encourage you to try to solve the third step by yourself, using some of the ideas we used in the 2nd step. Ask yourself :


*

*What do I want to show ? What to I need to find to reach the conclusion ?

*What does $I_{C_1} \cup I_{C_2} $ represent ?

*What set seems natural to introduce to reach our conclusion ? Can I define it as is or is there some subtleties ?

For reference, here is a full proof. Let $C_1$ and $C_2$ be subsets of $X^{k_1}$ and $X^{k_2}$. We want to show $I_{C_1} \cup I_{C_2} \in \A$, and therefore want to fing a set $C$ included in some $X^k$ such that $I_{C_1} \cup I_{C_2}  = I_C$.
$I_{C_1} \cup I_{C_2}$ represent all the sequence of $X^\N$ whose beginning is one of the finite sequence contained in $C_1$ of in $C_2$. Therefore, one set that seems natural for $C$ is to take $C= C_1 \cup C_2$. However, this is not directly possible as the $C_n$ are not subsets of the same set : $C_1$ is a subset of $X^{k_1}$ and $C_2$ is a subset of $X^{k_2}$. If $k_1=k_2$, then it works. : ideally, we would want to use the same $X^k$ for $I_{C_1} and I_{C_2}$.
Now, we can see that, given $k < l$ and $D \subset X^k$, we can build
$$D' = \{(x_1,x_2,\dots,x_k,\dots,x_l)\in X^l \mid (x_1,x_2,\dots,x_k) \in D\}.$$
Then the sequences of $X^{\N}$ whose beginning is in $D$ are exactly the same as  the sequences of $X^{\N}$ whose beginning is in $D'$. Therefore,
$$I_D = I_{D'}$$
even if the set $D'$ in a subset of $X^l$ and not $X^k$.
Thus we set $l = \max(k_1,k_2)$, and we define $C = (C_1)' \cup (C_2)' \subset X^l$. Then one can see (similar proof as in 2) the
$$I_{(C_1)'\cup(C_2)'} = I_{(C_1)'}\cup I_{(C_2)'} = I_{C_1}\cup I_{C_2}$$
which conclude our proof.
