$\sigma$-algebra generated by a subset of a set

Question

Let $\Omega$ = {1,2,3,4} and consider the collection of subsets of $\Omega$, $\mathcal A$={{1,2}}. The problem is to find the $\sigma$-algebra $\mathcal F$ generated by $\mathcal A$. I know this is a very basic question, but it is difficult for me to understand the concept of $\sigma$(X). If someone can explain the details of it, I will be greatly grateful.

Answer 1

From wiki "In mathematical analysis and in probability theory, a σ-algebra (also σ-field) on a set X is a collection Σ of subsets of X that includes the empty subset, is closed under complement, and is closed under countable unions and countable intersections."

Then your answer would be { \emptyset , {1,2}, {3,4} , {1234}}. By definition, it must contain the empty set, and to be closed under compliment that means it must contain the entire set. Also by adding {1,2} to the set, to keep the closure under compliment- we must include {3,4}.

Answer 2

The definition says that $\sigma(\mathcal{A})$ is the smallest $\sigma$-algebra on $\Omega$ which contains $\mathcal{A}$ (that is, contains all the subsets of $\Omega$ that are the elements of $\mathcal{A}$.

Before going on, try to answer: How many elements are there in $\mathcal{A}$ (that is, what is $\#\mathcal{A}$)? Is it $1$ or $2$?

Now remember that a $\sigma$-algebra must contain complements of each of its members and also must contain the union of any finite or countably infinite collection of them.

So we have that:

• $\{1,2\}$ must belong to $\sigma(\mathcal{A})$ (because it is an element of $\mathcal{A}$);
• $\Omega \setminus \{1,2\}=\{3,4\}$ must belong to $\sigma(\mathcal{A})$ (since it is a $\sigma$-algebra);
• And you must always have $\Omega$ itself, since it is the union of any set and its complement; and also the empty set $\emptyset$, which is the complement of $\Omega$.

So we have that $$\sigma(\mathcal{A})\supset \big\{\emptyset,\{1,2\},\{3,4\},\Omega\big\}.$$ But is it enough?

Well, try to prove or convince yourself that this is in fact a $\sigma$-algebra, and so we have a $\sigma$-algebra containing $\mathcal{A}=\big\{\{1,2\}\big\}$. But we've seen that we can't exclude any of those sets (that is, those elements of the $\sigma$-algebra): so there can't be a smaller $\sigma$-algebra containing $\mathcal{A}$. Then, actually $$\sigma(\mathcal{A}) =\big\{\emptyset,\{1,2\},\{3,4\},\Omega\big\}.$$