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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.

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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 $C_1\not\subset C_2$.

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  • $\begingroup$ 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 $\sigma(C_1) = \sigma(C_2) = \mathcal{P}(\Omega)$. $\endgroup$ – Viktor Glombik Mar 29 at 21:54
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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 $C_1 \not\subset C_2$.

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