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This is my homework exercise and I am stuck in following:

Exercise Let $\mathcal{A}_0$ be a countable family of subsets of a set $\Omega$. Show that the smallest set algebra $\mathcal{A}_{*}$ containing $\mathcal{A}_{o}$ is countable as well.

My idea is to show that the following set $\mathcal{A}_{*}$ is an algebra:

$$\mathcal{A}_{*}=\left ( \bigcup_{i=1}^{n} \bigcap_{j=1}^{m} A_{ij} : m, n \in \mathbb{N}, A_{ij} \in \mathcal{A}_{o} \cup \left \{ \emptyset, \Omega \right \} \cup \left \{A_{0}^{c} : A_{0} \in \mathcal{A}_{0} \right \}\right )$$

What I have done:

1) $\Omega \in \mathcal{A}_{*}$ for $m=n=1$ this is obvious.

2) $A=\bigcup_{i=1}^{n_1} \bigcap_{j=1}^{m_1} A_{ij}, \ A_{ij} \in \mathcal{\tilde{A}}_{o}:= \mathcal{A}_{o} \cup \left \{ \emptyset, \Omega \right \} \cup \left \{A_{0}^{c} : A_{0} \in \mathcal{A}_{0} \right \}$

$B=\bigcup_{i=1}^{n_2} \bigcap_{j=1}^{m_2} B_{ij}, \ B_{ij} \in \mathcal{\tilde{A}}_{o} \ \text{same as above} $

Now my problem is to express $A \cup B = \bigcup_{j=1}^{?} \bigcap_{i=1}^{?} C_{ij}$ what are the intervals of union and intersection to conclude that $A \cup B \in \mathcal{A}_{*}$.

3) The other problem is to show that $\mathcal{A}_{*}$ is closed under complements.

For this what I have done is: Suppose $A \in \mathcal{A}_{*}$, to show that $A^{c} \in \mathcal{A}_{*}$

$$A=\bigcup_{i=1}^{n_1} \bigcap_{j=1}^{m_1} A_{ij}, \ A_{ij} \in \mathcal{\tilde{A}}_{o}$$ where $A_{ij} \in \mathcal{\tilde{A}}_{o}$.

$$A^{c}=\left ( \bigcup_{i=1}^{n_1} \bigcap_{j=1}^{m_1} A_{ij}, \ A_{ij} \in \mathcal{\tilde{A}}_{o} \right )^{c}=\bigcap_{i=1}^{n_1} \bigcup_{j=1}^{m_1}A_{ij}^{c} \in \mathcal{A}_{*} ?? $$

EDIT It was a mistake for $A \cup B$ in the second step.

Melina
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1 Answers1

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For 1, we can assume that $n_1=n_2$ and $m_1=m_2$. Why? We can take dummy sets as an empty set or $\Omega$ appropriately. Then remainders are easy to handle.

For 2, the distribution law $$\bigcup_{i=1}^n\bigcap_{j=1}^m A_{i,j} = \bigcap_{f:[n]\to [m]}\bigcup_{i=1}^n A_{i,f(i)}$$ might be helpful. Here $[n]=\{1,2,\cdots,n\}$ and the intersection of the right-hand-side ranges over the set of functions from $[n]$ to $[m]$.

Hanul Jeon
  • 27,376
  • for $n_1=n_2$ and $m_1=m_2$ would $A \cup B = \bigcup_{i=1}^{n_1} \bigcap_{j=1}^{m_1} (A_{ij} \cup B_{ij})= \bigcup_{i=1}^{n_1} \bigcap_{j=1}^{m_1} C_{ij}$ – Melina May 03 '18 at 12:24
  • is this right?? – Melina May 03 '18 at 12:24
  • @Melina You need a distribution law to evaluate $A\cup B$. Notice that $$\left(\bigcap_{i=1}^m A_i \right)\cup\left(\bigcap_{j=1}^m B_j\right) = \bigcap_{(i,j)\in [m]\times[m]} (A_i\cup B_j).$$ – Hanul Jeon May 03 '18 at 22:12