Grimanijd

2022-07-06

Let $\mathcal{G}$ be any countable family of sets, where $\varphi \in \mathcal{G}$.
Let ${\mathcal{G}}_{1}$ be the family of all finite union of difference of sets in $\mathcal{G}$.
Let ${\mathcal{G}}_{2}$ be the family of all finite union of difference of sets in ${\mathcal{G}}_{1}$.
May be I am misunderstanding the meaning of "finite union of difference" but to me here ${\mathcal{G}}_{1}={\mathcal{G}}_{2}$. However my text book says it is not necessary. How so?
This is used in the textbook to prove the ring generated by countable generator is countable.

Charlee Gentry

Expert

Let $\mathcal{G}=\left\{\varnothing ,\left\{1\right\},\left\{2\right\},\left\{1,2,3,4\right\}\right\}$.
Then ${\mathcal{G}}_{1}=\left\{\varnothing ,\left\{1\right\},\left\{2\right\},\left\{1,2\right\},\left\{2,3,4\right\},\left\{1,3,4\right\},\left\{1,2,3,4\right\}\right\}$.
Now $\left\{3,4\right\}=\left\{2,3,4\right\}\setminus \left\{2\right\}$ is an element of ${\mathcal{G}}_{2}$, but $\left\{3,4\right\}\notin {\mathcal{G}}_{1}$.

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