Subsets of a set generating a sigma algebra

Question

I have a notation difficulty while writing correctly a sigma algebra generated not by a set of elements but rather by a set of subsets of a set.

Let $E=\{ 1,2,3,4\}$. I need to determine the $\sigma$-algebra generated by the family of subsets $X=\{\{1\},\{2\}\}$ of $E$.

I first thought about this solution: $\sigma(\{1\}, \{2\})=\{\emptyset ,\{1,2\}, \{3,4\}, \{1,2,3,4\}\}$ which seems not to hold.

I am wondering, whats the complement of $X$ in respect to $E$ ? Also, can one say that $\{\{1\},\{2\}\}$ is a subset of $E$ ?

Thanks for your comment.

Answer 1

First of all the generated sigma algebra must contain the sets that generate it. So obviously if we call the generated sigma algebra by $\Omega$ then $\{1\},\{2\}\in\Omega$. Also by definition $\emptyset, E\in\Omega$. Now we have to add complements and unions. So we get that $\Omega$ also contains the following sets: $\{2,3,4\},\{1,3,4\},\{1,2\},\{3,4\}$. And that is all, you can check the elements of $\Omega$ are closed under complement and union, so $\Omega=\{\emptyset,\{1\},\{2\},\{1,2\},\{3,4\},\{2,3,4\},\{1,3,4\},\{1,2,3,4\}\}$.

As for your last question, obviously $\{\{1\},\{2\}\}$ is not a subset of $E$. It is a subset of $2^E$, the power set. And by the way a sigma algebra on $E$ contains elements of $2^E$, not of $E$ itself.

Answer 2

By the definition of $\sigma$-algebra generated by a set (i.e. smallest $\sigma$-algebra containing given set),

$\sigma(X) = \{ \emptyset, \{1\},\{2\},\{1,2\},\{3,4\},\{2,3,4\},\{1,3,4\},E\}$

You can check that above collection satisfies all properties required for $\sigma$-algebra. It contains $X$ and it’s smallest $\sigma$-algebra containing $X$.

Answer 3

If $X=\{\{1\},\{2\}\}$ then $\sigma(X)$ is by definition the smallest $\sigma$-algebra such that $X\subseteq\sigma(X)$.

So $\sigma(X)$ must "look like" $\{\{1\},\{2\},\dots\}$ here, showing that your solution is not correct.

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If $X=\{A_1,\dots,A_n\}$ where the $A_i$ are distinct subsets of $E$ then $\sigma(X)$ will contain the collection:$$\mathcal V=\{B_1\cap\cdots\cap B_n\mid B_i\in\{A_i,A_i^{\complement}\}\text{ for }i=1,\dots,n\}$$

Note that this collection has at most $2^n$ distinct element, that two distinct elements that belong to $\mathcal V$ are disjoint and finally that $\mathcal V$ is a cover of $E$.

Then $\sigma(X)$ will be exactly the collection of unions of elements of $\mathcal V$.

There will be at most $2^{2^n}$.

Applying this you can find $\sigma(X)$ in your case (good exercise!)