Prove that $\sigma (\cap_{i \in I} C_i)=\cap_{i \in I} \sigma (C_i)$

Question

Do we have the following identity?

$$\sigma (\cap_{i \in I} C_i)=\cap_{i \in I} \sigma (C_i)$$

Here $C_i$ is a subset of a set $\Omega$.

Answer 1

No.

Let $\Omega = [0,1]$. Let $C_1 = [1/4,1/2]$ and let $C_2 = [1/3,2/3]$. Then $$ \sigma(C_1) = \{\emptyset, [1/4,1/2], [0,1/4) \cup (1/2,1], [0,1]\}$$ and $$ \sigma(C_2) = \{\emptyset, [1/3,2/3], [0,1/3) \cup (2/3,1], [0,1]\}$$ so that $\sigma(C_1) \cap \sigma (C_2) = \{\emptyset, [0,1]\}$.

On the other hand, $C_1 \cap C_2 = [1/3,1/2]$ so that $$ \sigma (C_1 \cap C_2) = \{\emptyset, [1/3,1/2], [0,1/3) \cup (1/2,1], [0,1]\}.$$