Reading through my lecture notes, and I'm stuck a bit on this concept.
Let $A$ be a set of subsets of $E$. Define $$ \sigma(A) = \{ A \subseteq E \ : \ A \in F \text{ for all } \sigma\text{-algebras } F \text{ containing }A \} .$$ Then $\sigma(A)$ is a $\sigma$-algebra, which is called the $\sigma$-algebra generated by $A$. It is the smallest $\sigma$-algebra containing $A$.
So let's say: $E = \{1,2,3\}$. All the possible subsets of $E$ will be $ = \{ \{ \emptyset \}, \{1,2,3\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\} \}$. So if we choose $A= \{1 \}$, then what would $\sigma(\{1\})$ equals to?
$$ \sigma(\{1\}) = \{ \{1 \} \subseteq E \ \colon \ \{1 \} \in F \text{ for all } \sigma\text{-algebras } F \text{ containing } \{1\} \} .$$
How would I find all the $\sigma$-algebras $F$ containing $\{1\}$? Thanks.
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$\begingroup$As a general comment on this kind of construction, I might suggest reading this answer.
Generally speaking, what you describe is the "top-down" approach to the construction. It is not, generally, practical to actually know the description of the elements in the generated object. For that, you want the "bottoms-up" construction. In the question linked-to above, Asaf Kargila provides the description of the bottoms-up construction.
In your particular case, of course, since $E$ is small (3 elements), so $P(E)$ is small (8 elements), the number of possible $\sigma$-algebras is somewhat manageable (though still large).
What are all $\sigma$-algebras on $E=\{1,2,3\}$ that contain $\{1\}$ as an element? They must contain $\emptyset$, $\{1\}$, $\{2,3\}$, and $E$. There are 4 other elements in $P(E)$ which may or may not be in a $\sigma$-algebra.
- One $\sigma$-algebra is just $\{\emptyset, \{1\}, \{2,3\}, E\}$.
- If the $\sigma$-algebra contains either $\{2\}$ or $\{3\}$ in addition to those, then it must also contain the other (symmetric difference with $\{2,3\}$), and hence be all of $P(E)$.
- If the $\sigma$-algebra contains any other $2$-element subset, then it must contain another singleton (symmetric difference with $\{2,3\}$), hence must be all of $P(E)$.
So in fact the only $\sigma$-algebras on $E$ that contain $\{1\}$ are $\{\emptyset, \{1\},\{2,3\}, E\}$ and $P(E)$. The intersection is just $\{\emptyset,\{1\},\{2,3\},E\}$, so the $\sigma$-algebra generated by $\{1\}$ is $\{\emptyset, \{1\}, \{2,3\}, E\}$.
$\endgroup$ 1 $\begingroup$Note that the correct definition is what Martin has written. The definition of $\sigma(A)$ in general is not very useful to construction of the $\sigma$-algebra generated by $A$, because as you have mentioned it may be hard to describe all $\sigma$-algebras which contain $A$, but when $A$ is just a single set the problem is easy.
Let $P(E)$ denote the powerset of $E$, i.e. the set of all subsets of $E$ and $a\in P(E)$ - subset of $E$. Let us consider $F:=\sigma(a)$.
First of all, $E\in F$ be definition;
$\emptyset\in F$ and $a^c\in F$ because any complement of element of $F$ belongs to $F$.
Let us assume now that $F = \{\emptyset,a,a^c,E\}$ since we prove that these elements have to be in $F$ anyway. We need to check that $E\in F$ (done), $F$ is closed under taking complements (done) and any countable union of elements of $F$ belongs to $F$ (also easy to see).
As a result, $\sigma(a) = \{\emptyset,a,a^c,E\}$. In your case $a = \{1\}$ hence $$ \sigma(\{1\}) = \{\emptyset,\{1\},\{2,3\},\{1,2,3\}\}. $$
$\endgroup$ 1 $\begingroup$Actually $A=\{\{1\}\}$ since it must be a subset of the Power set. Now to find $\sigma(A)$, I would do the following:
First $\emptyset, E, \{1\}$ must be in $\sigma(A)$ and then the complement of $\{1\}$ which is $\{2,3\}$. So that$$ \sigma(A)=\{\emptyset, E, \{1\}, \{2,3\} \} $$
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