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I have a simple questions concerning how $\sigma$-Algebra and Algebra are defined in my textbook. In it, it defines:

$\sigma$-Algebra is a class of sets $A$ st:

  1. $\Omega \in A$ (omega being the entire space)
  2. A is closed under set complements
  3. A is closed under countable unions

And Algebra is defined as a class st:

  1. $\Omega \in A$ (omega being the entire space)
  2. A is closed under set Differences
  3. A is closed under unions

My confusion comes from the set difference vs set complements part. It seems to me that closure under set difference is a stronger condition than closure under set complements. But the textbook seems to be treating a $\sigma$-Algebra as just an algebra that is closed under countable unions. Are set difference and complements equivalent due to the other conditions?

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    $\begingroup$ They are they same, except an algebra is closed under FINITE unions, whereas a $\sigma$-algebra is closed under COUNTABLE unions. $\endgroup$
    – Dzoooks
    Commented Jun 5, 2018 at 18:33
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    $\begingroup$ Your textbook's definitions are strange. Both algebras and $\sigma$-algebras are closed under complements and set differences. Usually the definition is stated in terms of closure under complements. I've only seen "closed under set differences" in the definition of a $\sigma$-ring, which doesn't necessarily contain $\Omega$. $\endgroup$
    – user169852
    Commented Jun 5, 2018 at 18:35
  • $\begingroup$ I think it was trying to show us multiple ways to confirm the same conditions (for example it uses de-morgan's to interchangeably use unions and intersections, but i think tends to make you think that they are totally separate objects) $\endgroup$
    – Anonymous Emu
    Commented Jun 5, 2018 at 18:37

1 Answer 1

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Any $\sigma$-algebra is closed under set differences.

This is because $A\backslash B= A\cap (B^c$).

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  • $\begingroup$ For the record, what about the converse? Is it sufficeint to prove that a class is an algebra by proving that it is closed under complements? Or does it have to be set difference? $\endgroup$
    – Anonymous Emu
    Commented Jun 5, 2018 at 18:35
  • $\begingroup$ Do you mean $B^c= X\backslash B$? so close under set differences does imply that it is closed under complements. $\endgroup$
    – Yanko
    Commented Jun 5, 2018 at 18:36
  • $\begingroup$ Oh i get that closure under set difference imples closure under complements. And in general the other direction is not true. But If i want to prove that something is an algebra, if I have 1) $X \in A$ and closed under unions, is it sufficient to prove that it is closed under complements? or do I have to prove that it is closed under set differences? $\endgroup$
    – Anonymous Emu
    Commented Jun 5, 2018 at 18:39
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    $\begingroup$ Yes it is enough. Take $A,B$ in your algebra, then $A^c,B$ in your algebra, since it is closed under (finite) union $A^c\cup B$ is in your algebra. Take complement again you have that $A\cap B^c = A\backslash B$ is in your algebra. $\endgroup$
    – Yanko
    Commented Jun 5, 2018 at 18:40
  • $\begingroup$ Thank you so much. $\endgroup$
    – Anonymous Emu
    Commented Jun 5, 2018 at 18:41

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