# Prove that $\mathcal{A}$ is an algebra

Let $$X=\{1,\ldots,m\}$$ and

$$$$r:X\rightarrow [0,1] \\ i\mapsto r_i^s$$$$

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 $$$$I_{C}=\{(x_1,x_2,\ldots): (x_1,x_2,\dots,x_k)\in C\}.$$$$

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 $$I_m\cup I_k\in \mathcal{A}$$?

New contributor
habit is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

$$\def\N{\mathbb{N}}$$ $$\def\A{\mathcal{A}}$$

The sets $$I_C$$ are a subset of $$X^{\N}$$. You thus want to prove

1. $$X^{\N} \in \A$$ ;
2. Given $$C \subset I_k$$, $$X^{\N} \setminus I_C \in \A$$ ;
3. 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 !

1. For $$C = X \subset X^1$$, we have $$I_C = X^{\N}$$, therefore $$X^{\N} \in \A$$.
2. 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.

1. 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.