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This is adapted from 1.7.7 in Friedman's "Foundations of Modern Analysis":

Let $\mathscr{B}$ be the $\sigma$-ring generated by the class of open subsets of $X$ [a fixed set], and $\mathscr{D}$ the $\sigma$-ring generated by the class of closed subsets of $X$. Show that $\mathscr{D} = \mathscr{B}$.

I would appreciate a hint on how to begin doing this exercise, because I haven't a clue. I'm not really sure what the problem entails. I know what a $\sigma$-ring is, as well as open and closed sets. That $\mathscr{B},\mathscr{D}$ are generated means that they each are, in a sense, the smallest and unique ring containing its respective "underlying" class of sets. But none of this gives me any idea on where to start. The problem is from a section on metric spaces (prior to metric outer measures), but again not even the context gives me any ideas.

(Couldn't come up with a good title, change it if you like...)

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Hint: show that every closed set is in $\mathscr B$ and every open set is in $\mathscr D$.

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Thank you. Two problems: How, and more importantly perhaps, why does this show that the $\sigma$-rings are equal? Anyway I don't know how to show what you say. What are the legal operations? I don't get this... – Erik Vesterlund Oct 15 '13 at 14:36
$\mathscr D$ is the intersection of all $\sigma$-rings containing the closed sets. If $\mathscr B$ is a $\sigma$-ring that contains the closed sets, that implies $\mathscr D \subseteq \mathscr B$. Similarly in the other direction. – Robert Israel Oct 15 '13 at 15:09
Look at the definition of $\sigma$-ring. How are closed sets related to open sets? – Robert Israel Oct 15 '13 at 15:13
I'm looking intensely at the definition but fail to make the relevant connection. The complement of an open set is a closed set, is that what you mean? If so I'm still lost, complementation is not in the definition of $\sigma$-rings. – Erik Vesterlund Oct 15 '13 at 23:18
Relative complementation is in the definition. Recall that $X$ itself is both open and closed in $X$. – Robert Israel Oct 16 '13 at 2:08

you can try to show that for a set $b \in \mathcal B $, b also belongs to $\mathcal D$

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