Sigma-algebra of two subsets

Question

Let $X$ be a set and $A, B \subset X$. Define $\varepsilon:=$ {A,B} $\in$ $\mathcal{P}(X)$ .

How to find the sigma-algebra generated by $\varepsilon$?

I got $\sigma(\varepsilon)$=
{$\emptyset, X, A, A^c, B, B^c, A \cup B, A^c \cup B, A \cup B^c, A^c \cap B^c, A \cap B, A^c \cup B, A \cap B^c, A^c \cap B^c$}.

Besides, how many elements can $\sigma(\varepsilon)$ have at most? I got 14 (elements of the sigma-algebra): $\emptyset, X, A, A^c, B, B^c, A \cup B, A^c \cup B, A \cup B^c, A^c \cap B^c, A \cap B, A^c \cup B, A \cap B^c, A^c \cap B^c.$

Also, how to choose examples of $X, A, B$ such that $\sigma(\varepsilon)$ has less elements than the maximum?

Answer 1

More generally let $A_1,\dots,A_n$ be subsets of $X$.

These sets induce the following "pseudo"-partition of $X$:$$\mathcal P=\{E_1\cap\cdots\cap E_n\mid E_i\in\{A_i,A_i^{\complement}\}\text{ for }i=1,\dots,n\}$$ Observe that every pair of distinct elements of $\mathcal P$ has an empty intersection and that $X$ is covered by the elements of $\mathcal P$. It is not necessarily a partition because $\mathcal P$ might contain the empty set.

Further observe that $\mathcal P$ has at most $2^n$ elements. If $\mathcal A=\sigma(\{A_1,\cdots,A_n\})$ denotes the $\sigma$-algebra generated by the sets $A_1,\dots,A_n$ then evidently $\mathcal P\subseteq\mathcal A$.

Now define:$$\mathcal A'=\{\cup\mathcal V\mid\mathcal V\subseteq\mathcal P\}$$in word: $A$ is an element of $\mathcal A'$ if and only if it can be written as a union of elements of $\mathcal P$.

Since $\mathcal P$ is finite every element of $\mathcal A'$ will be a finite union of elements of $\mathcal P$ so that from $\mathcal P\subseteq\mathcal A$ we can conclude that also $\mathcal A'\subseteq\mathcal A$.

Secondly it is easy to see that $\mathcal A'$ is closed under arbitrary unions (a union of unions is again a union).

Also it is not difficult to see that $\mathcal A'$ is closed under complements.

For that just note that $(\cup\mathcal V)^{\complement}=\cup(\mathcal P-\mathcal V)$.

This tells us that $\mathcal A'$ is a $\sigma$-algebra.

Finally observe that $A_i\in\mathcal P$ for every $i\in\{1,\dots\}$.

As an example note that: $$A_1=\cup\{A_1\cap E_2\cap\cdots\cap E_n\mid E_i\in\{A_i,A_i^{\complement}\}\text{ for }i=2,\dots,n\}$$

These facts together tell us that $\mathcal A'=\mathcal A$.

On base of the fact that $\mathcal P$ has at most $2^n$ elements we find on base of the definition of $\mathcal A'$ that $\mathcal A=\mathcal A'$ has at most $2^{2^n}$ elements (any element of $\mathcal P$ belongs to the union or belongs not to the union).

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Your $14$ is wrong and should be $2^{2^2}=2^4=16$.

The sets:

• $\varnothing$

(empty union)

• $A\cap B$
• $A\cap B^{\complement}$
• $A^{\complement}\cap B$
• $A^{\complement}\cap B^{\complement}$

(exactly $1$ element of $\mathcal P$ in the union)

• $A$
• $B$
• $A^{\complement}$
• $B^{\complement}$
• $(A\cap B)\cup(A^{\complement}\cap B^{\complement})$
• $(A^{\complement}\cap B)\cup(A\cap B^{\complement})$

(exactly $2$ elements of $\mathcal P$ in the union)

• $X-(A\cap B)=A^{\complement}\cup B^{\complement}$
• $X-(A\cap B^{\complement})=A^{\complement}\cup B$
• $X-(A^{\complement}\cap B)=A\cup B^{\complement}$
• $X-(A^{\complement}\cap B^{\complement})=A\cup B$

(exactly $3$ elements of $\mathcal P$ in the union)

• $X$

(exactly $4$ elements of $\mathcal P$ in the union)

Answer 2

If $A,B$ are "most general" (not $A \subseteq B$ or reversely, non-empty intersection and $A \cup B \neq X$) we get in the $\sigma$-algebra all sets $ A \cap B, A \setminus B, B \setminus A, X\setminus (A \cup B)$, which are 4 (non-empty, as I assumed) sets that form a disjoint partition of $X$. So if we just add all possible unions of those $4$ we are done generating the $\sigma$-algebra, because we cannot add any more by intersections (except $\emptyset$ which we already get as the empty union, i.e. the union of $0$ sets from the partition, anyway) and complements of unions are unions already (in a partition). So we get $2^4$ sets in the generated $\sigma$-algebra (or just algebra, as we're never going to get infinite unions or intersections anyway, starting from finitely many sets).

As a result I get $$\{\emptyset, A^c \cap B^c, A^c \cap B, A^c, A \cap B, (A \Delta B)^c, B, A^c \cup B, A \cap B^c, B^c, A \Delta B, A^c \cup B^c, A, B^c \cup A, A \cup B, X\}$$