# Finding the $\sigma$-algebra generated by $\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}$.

In my study of $$\sigma$$-algebra, I am having difficulty understanding the concept of the smallest $$\sigma$$-algebra.

Here, the set $$\mathcal{F}\in P(\Omega)$$ (where $$P(\Omega)$$ denotes the power set of $$\Omega$$) is a $$\sigma$$-algebra if $$\mathcal{F}$$ has the following properties: \begin{align} (1):& \ \ \emptyset\in\mathcal{F} \\ (2):& \ \ A\in\mathcal{F}\implies A^c\in\mathcal{F} \\ (3):& \ \ A_1,A_2,...\in\mathcal{F}\implies \bigcup_{i=1}^\infty A_i\in\mathcal{F} \end{align}

Now, given a family $$\mathcal{U}$$ of subsets of $$\Omega$$, there is a smallest $$\sigma$$-algebra $$\sigma_{\mathcal{U}}$$ containing $$\mathcal{U}$$. This $$\sigma$$-algebra is the intersection of all $$\sigma$$-algebra containing you.

To help improve my understanding, I considered the following example. Let $$\Omega=\{1,2,3,4,5\}$$ and let $$\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}=\{A,B\}$$. Then,

\begin{align} \sigma_{\mathcal{U}}&=\{\underbrace{\emptyset}_{\text{by (1)}},\underbrace{\Omega,A,B,A^c,B^c}_{\text{by (2)}},\underbrace{A\cup B,A^c\cup B, A\cup B^c, A^c\cup B^c}_\text{by (3)},\underbrace{(A\cup B)^c,(A^c\cup B)^c,(A\cup B^c)^c,(A^c\cup B^c)^c}_\text{by (2)}\} \\ &=\{\emptyset,\{1,2,3,4,5\},\{1,2,3\},\{3,4,5\},\{4,5\},\{1,2\}\} \end{align}

Is this correct?

## 1 Answer

Assuming that $$A$$ equals $$\{1,2,3\}$$, and not $$\{1,2\}$$, the sigma-algebra generated by $$\{A, B\}$$ is the power set of the three sets $$\{1,2\}$$, $$\{3\}$$, $$\{4,5\}$$, hence it has eight elements. (You are missing the set $$\{3\}$$ and its complement $$\{1,2,4,5\}$$.)

This follows from the result: If the universe $$\Omega$$ is countable, then every sigma-algebra on $$\Omega$$ can be generated by a partition of $$\Omega$$.

So to determine $$\sigma({\mathcal U})$$ for your example, your first step is to find the partition that corresponds to $$\mathcal U$$: that is, for every $$\omega\in\Omega$$ find the smallest member of $$\sigma({\mathcal U})$$ that contains $$\omega$$. Check that there is no way to separate $$1$$ and $$2$$ using unions and/or intersections from $$\mathcal U$$, and similarly you cannot find a way to separate $$4$$ from $$5$$. On the other hand, the element $$3$$ is covered by the singleton set $$\{3\} = \{1,2,3\}\cap\{3,4,5\}$$. The three sets $$\{1,2\}$$, $$\{3\}$$, $$\{4,5\}$$ are disjoint and their union is $$\Omega$$, so that's your partition.

• Does the set $\{3\}$ correspond to any of my terms given by $\sigma_{\mathcal{U}}=\{\cdots\}$? – M B Mar 10 at 6:20
• Yes, $\{3\} = A\cap B = (A^c\cup B^c)^c$. – grand_chat Mar 10 at 6:23