I'm currently reading through some set theory questions for a statistics class and I came across the following notation that I've never seen before, and I can't seem to find anywhere else online. The equation in question is: $$ \mathcal{B} = \cap\{\mathcal{E}:\mathcal{D}\subset\mathcal{E} \text{ and } \mathcal{E} \text{ is a } \sigma \text{-field}\} $$ My question is, what does the intersection symbol in front of the set builder notation mean? How can I take the intersection of only one set? I would ask my professor, but they don't have any office hours today and I'd rather not wait too long for what seems like a simple question.

  • 1
    $\begingroup$ Probably the intersection of all the valid $\mathcal E$. $\endgroup$
    – player3236
    Sep 28, 2020 at 18:57

1 Answer 1


The notation $\cap A$ usually denotes the intersection of all the sets that are contained in $A$, although not everyone uses this notation. Using the more common notion of an intersection over an index set, this means that

$$ \cap A = \bigcap_{B\in A} B $$ holds, and that $A$ is a set of sets.

In your case, this means $$ \mathcal{B} = \cap\{\mathcal{E}:\mathcal{D}\subset\mathcal{E} \text{ and } \mathcal{E} \text{ is a } \sigma \text{-field}\} = \bigcap_{\mathcal{E}:\mathcal{D}\subset\mathcal{E} \text{ and } \mathcal{E} \text{ is a } \sigma \text{-field}} \mathcal{E} $$

  • $\begingroup$ So in this case, the result would be the smallest such $\mathcal{E}$? $\endgroup$ Sep 28, 2020 at 19:07
  • $\begingroup$ @GideonTveten yes $\endgroup$
    – supinf
    Sep 28, 2020 at 19:07
  • $\begingroup$ Thanks, that makes a lot more sense. $\endgroup$ Sep 28, 2020 at 19:08

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