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So I'm reading Folland and on p. 22 he says

"If X is any metric space..., the sigma algebra generated by the family of open sets in X is called the Borel sigma algebra and is denoted Bx."

He just went over how a subset of P(X) generates a sigma algebra but I don't understand how a family of open sets in X can generate a sigma algebra.

If we have a subset E of P(X), the sigma algebra generated by E is the smallest sigma algebra containing E. I don't quite see how this extends to a family of subsets of P(X). Should I just assume that, when multiple sets generate a sigma algebra, this sigma algebra is the smallest one containing all of the generating subsets?

Thanks for the help!

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2 Answers 2

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A subset E of P(X) is a family of subsets of X, not a family of subsets of P(X).

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  • $\begingroup$ Say X = {1, 2, ... , 10}. Then a subset E of P(X) could be {{1,2}, {3,9}}. Whereas a family of subsets of P(X) could be {{{1,2}, {3,9}}, {{1,2}, {7,8}}}. Yes? $\endgroup$
    – anon_swe
    Commented Dec 17, 2013 at 22:25
  • $\begingroup$ Exactly. $ $ $ $ $\endgroup$
    – Did
    Commented Dec 17, 2013 at 22:27
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Let $E:=\{$open sets in $X\}$. Then we have $E\subseteq P(X)$. All is fine.

So, the Borel sigma algebra in particular contains all open sets, all countable intersections of open sets, all closed sets, all countable unions of closed sets, all countable intersections of these, and so on...

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  • $\begingroup$ I agree with that, but are you saying I should assume "when multiple sets generate a sigma algebra, this sigma algebra is the smallest one containing all of the generating subsets?" $\endgroup$
    – anon_swe
    Commented Dec 17, 2013 at 20:35
  • $\begingroup$ Yes. I don't really understand where your confusion comes from. $E$ is one subset of $P(X)$, and it generates the smallest sigma algebra which contains $E$, as you once understood within the body of the question. $\endgroup$
    – Berci
    Commented Dec 17, 2013 at 20:37
  • $\begingroup$ OK just clarifying. Thanks! $\endgroup$
    – anon_swe
    Commented Dec 17, 2013 at 20:42

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