# $\sigma(C_1) \subset \sigma(C_2) \iff C_1 \subset C_2$?

Let $$(\Omega, \mathcal{F})$$ be a measurable space and $$C_1, C_2 \subset \mathcal{P}(\Omega)$$. Then, we know that $$C_1 \subset C_2 \implies \sigma(C_1) \subset \sigma(C_2)$$, where $$\sigma(A) := \bigcap \{ \mathcal{B}: \mathcal{B} \text{ is a } \sigma\text{-algebra containing } A \}$$ is the generated $$\sigma$$-algebra of $$A \subset \mathcal{P}(\Omega)$$. Does the converse also hold?

Since in our lecture, we only formulated one direction, I would be surprised if both directions worked, but haven't been able to come up with a counterexample.

Any help is greatly appreciated.

I assume here your "$$\subset$$" means "$$\subseteq$$" (containment) and not "$$\subsetneq$$" (proper containment).
Assuming this, then it's not true. Suppose $$S$$ is an element of $$\sigma(C_2)$$ which is not actually already in $$C_2$$, and define $$C_1 = C_2\cup \{S\}$$.
Then $$\sigma(C_1)=\sigma(C_2)$$, so $$\sigma(C_1)\subset \sigma(C_2)$$, but $$C_1\not\subset C_2$$.
• Is the following a good example? $\Omega := \{ 0,1,2\}$, $C_1 := \{ \{1\}, \{2\}, \{0\} \}$ and $C_2 := \{ \{1,2\}, \{0,2\}, \{0,1\}\}$? Then, clearly, $C_1 \not\subset C_2$, but $\sigma(C_1) = \sigma(C_2) = \mathcal{P}(\Omega)$. – Viktor Glombik Mar 29 at 21:54
Take $$\Omega := \mathbb{R}, \ C_1 := \{ A \subset \Omega: A \text{ open} \}$$ and $$C_2 := \{ A \subset \Omega: A \text{ closed} \}$$. Then, $$\sigma(C_1) = \sigma(C_2) = \mathcal{B}$$, where $$\mathcal{B}$$ is the Borel-$$\sigma$$-Algebra on $$\mathbb{R}$$, and therefore, $$\sigma(C_1) \subset \sigma(C_2)$$ but $$C_1 \not\subset C_2$$.