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I started studying Probability, and I'm not sure if I understand what is the meaning of "closed under union".

A collection (say $F$) of subsets of a set (say $\Omega$) is said to be a $\sigma$-algebra if:

  • $\Omega \in F$
  • $F$ is closed under complement
  • $F$ is closed under union

Now, consider the following example:

Given $\Omega = \{1, 2, 3, 4, 5\}$, is $F = \{\emptyset, \{1, 2\}, \{3, 4, 5\}, \{5\}, \{1, 2, 3, 4\}, \{1, 2, 3, 4, 5\}\}$ closed under union?

I'm asking because I know that $\bigcup_{i = 1}^{6} X_i \in F$; however it does not happen for any two elements, for example: $\{1,2\} \cup \{5\} \not\in F$.

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    $\begingroup$ $\sigma$-algebras are only required to be closed under countable unions. $\endgroup$ Jan 6 '19 at 8:42
  • $\begingroup$ slight change to title; a collection of sets can be closed under union, not the set itself; $\endgroup$
    – Chris
    Jan 6 '19 at 9:09
  • $\begingroup$ You're right! Thanks, I changed it. $\endgroup$
    – bbublue
    Jan 6 '19 at 9:10
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 "$F$ is closed under union" means that for all $A,B \in F$, $A \cup B \in F$.

So here $\{1,2\}$ and $\{5\}$ are elements of $F$, but their union is not, so $F$ is not closed by union.

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“Closed under union” means that the union of any set of members of $F$ is also a member of $F$.

In your example $F$ is not closed under union (although it is closed under complement).

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