When determining the smallest sigma-algebra generated by a finite collection of sets (and hence the smallest algebra containing that collection), is there any faster way to do this than by direct computation? On a related note, is there a criterion for selecting a set which generates the finest sigma-algebra (that is, the power set)?
[Math] Sigma algebra generated by a set
measure-theory
Related Solutions
For Q1 and Q2: Yes, you are right. Actually, every countably generated $\sigma$-algebra is induced by a real valued function, so it comes from a separable pseudometrizable space.
For Q3: Yes, the Borel $\sigma$-algebra on $\mathbb{R}$ is a counterexample. You can add a countable number of points to any topology generating the usual Borel sets without changing the generated $\sigma$-algebra.
For Q4: The $\sigma$-algbera consisting of countable sets and sets with countable complements on $\mathbb{R}$ is a sub-$\sigma$-algebra of the Borel $\sigma$-algebra and is generated by the cofinite topology, which is not second countable. Since the countable-cocountable $\sigma$-algebra is also not countably generated, it doesn't come from any second countable topological space.
Let it be that $\langle\Omega,\mathcal F\rangle$ is a measurable space and that $f:\Omega'\to\Omega$ denotes a function.
Moreover let it be that $\mathcal C\subseteq\mathcal F$ such that $\mathcal F=\sigma(\mathcal C)$.
Then it can be shown that:$$f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C))\tag1$$
In the special case where $\Omega'\subseteq\Omega$ and $f$ denotes the inclusion application of $(1)$ results in:$$\{\Omega'\cap F\mid F\in\mathcal F\}=\sigma\left(\{\Omega'\cap C\mid C\in\mathcal C\}\right)$$ which is exactly what you are after.
Familiarity with $(1)$ is in my view a "must" in measure theory.
For a proof of it have a look at this answer.
edit:
Let $\Omega'\subseteq\Omega$ and let's denote $\mathcal{C}'=\left\{ A\cap\Omega'\mid A\in\mathcal{C}\right\} $ and $\mathcal{F}'=\left\{ A\cap\Omega'\mid A\in\mathcal{F}\right\} $ where $\mathcal{C}\subseteq\wp\left(\Omega\right)$ and $\mathcal{F}=\sigma\left(\mathcal{C}\right)$.
It is not difficult to prove that $\mathcal{F}'$ is a $\sigma$-algebra, and this with $\mathcal{C}'\subseteq\mathcal{F}'$. This allows the conclusion $\sigma\left(\mathcal{C}'\right)\subseteq\mathcal{F}'$ and this part you had done yourself allready.
Now let $\mathcal{A}:=\left\{ A\in\wp\left(\Omega\right)\mid A\cap\Omega'\in\sigma\left(\mathcal{C}'\right)\right\} $.
Again it is not difficult to prove that $\mathcal{A}$ is a $\sigma$-algebra, and this with $\mathcal{C}\subseteq\mathcal{A}$. This allows the conclusion $\mathcal{F}=\sigma\left(\mathcal{C}\right)\subseteq\mathcal{A}$. That means that $A\cap\Omega'\in\sigma\left(\mathcal{C}'\right)$ for every $A\in\mathcal{F}$ and states that $\mathcal{F}'\subseteq\sigma\left(\mathcal{C}'\right)$.
If $f:\Omega'\to\Omega$ denotes the inclusion then $\mathcal C'=f^{-1}(\mathcal C)$ and $\mathcal F'=f^{-1}(\mathcal F)=f^{-1}(\sigma(\mathcal C))$ so in the first part it is shown that $\sigma(f^{-1}(\mathcal C))\subseteq f^{-1}(\sigma(\mathcal C))$ and in the second part that $f^{-1}(\sigma(\mathcal C))\subseteq\sigma(f^{-1}(\mathcal C))$.
For completeness: $$f^{-1}(\mathcal C):=\{f^{-1}(A)\mid A\in\mathcal C\}=\{\{\omega\in\Omega'\mid f(\omega)\in A\}\mid A\in\mathcal C\}\tag2$$
So if $f:\Omega'\to\Omega$ is prescribed by $\omega\mapsto\omega$ (i.e. inclusion) then we can expand $(2)$ with:$$=\{\{\omega\in\Omega'\mid \omega\in A\}\mid C\in\mathcal C\}=\{A\cap\Omega'\mid A\in \mathcal C\}=\mathcal C'$$
Best Answer
Consider set $E_1,\dots,E_N$. Take $S_I:=\bigcap_{i\in I}E_i\cap\bigcap_{i\in [N]\setminus I}E_k^c$ for $I\subset [N]$. These sets are pairwise disjoint and generate the same $\sigma$-algebra as $E_1,\dots,E_N$. For each $I\subset [2^N]$, define $F_I:=\bigcup_{i\in I}S_i$. Then the $\sigma$-algebra generated by the $E_j$ is $\{F_I,I\subset [N+1]\}$.
Note that the $\sigma$-algebra generated by a finite collection is finite.