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Let $\mathcal{C}$ be any nonempty family of subsets of $\Omega$. Consider the collection of all $\sigma$-algebras that contain $\mathcal{C}$ ($\mathcal{C}\subset 2^\Omega$ for example). Let $\mathcal{O}$ be the intersection of all such $\sigma$-algebras. Show that $\mathcal{O}$ is a $\sigma$-algebra, the "smallest" $\sigma$-algebra, that contains $\mathcal{C}$. $\mathcal{O}$ is said to be the $\sigma$-algebra generated by $\mathcal{C}$.

I believe I should use the following conclusion but I am unsure how:

$\Omega$ is any space, $\mathcal{O}_1$ and $\mathcal{O}_2$ are each $\sigma$-algebras of subsets of $\Omega$. Let $\mathcal{O}_3$ be the collection of sets that belong to both $\mathcal{O}_1$ and $\mathcal{O}_2$, the so-called intersection of $\mathcal{O}_1$ and $\mathcal{O}_2.$ Show that $\mathcal{O}_3$ is a $\sigma$-algebra of $\Omega$.

Any suggestions on the proof?

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    $\begingroup$ You merely have to check that $\mathscr{O}$ satisfies the three conditions required for it to be a $\sigma$-algebra. This is straightforward. $\endgroup$ – BrianO Feb 2 '16 at 3:01
  • $\begingroup$ But does the second statement show that? $\endgroup$ – Username Unknown Feb 2 '16 at 3:06
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    $\begingroup$ "second statement" = what? Your 2nd to last paragraph? If that's what you mean, then no, of course it doesn't show that. That statement is the result you're trying to prove for the special case where $\mathscr{C}$ has one or two members. But you can mimic the proof of that. $\endgroup$ – BrianO Feb 2 '16 at 3:18

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